Three Dimensional Geometry

$$\begin{gathered} \frac{x-1}{1}=\frac{y-4}{3}-\frac{z-7}{5}=\lambda \left(l\mathrm{et}\right) \\ B(\lambda +1,3\lambda +4,5\lambda +7) \\ DR\text{'}s \text{of} AB=⟨\lambda ,3\lambda +6,5\lambda +8⟩ \\ Now, \lambda +9\lambda +18+25\lambda +40=0 \\ \lambda =\frac{-58}{35} \\ B=(\frac{-58}{35}+1,3\left(\frac{-58}{35}\right)+4,5\left(\frac{-58}{35}\right)+7) \\ =\left(\frac{-23}{35},\frac{-34}{35},\frac{-45}{35}\right)=\left(\alpha ,\beta ,\gamma \right) \\ \mathrm{Now}, \alpha +\beta +\gamma =\frac{-23-34-45}{35}=\frac{-102}{35} \end{gathered}$$

Let the point of intersection be \(A\)

\( \therefore A(1+\lambda a, 1-\lambda,-1) \equiv(4+2 \mu, 0,-1+\mu a) \)
\( \therefore 1-\lambda=0 \Rightarrow \lambda=1 \)
\( \mu a=0 \Rightarrow \mu=0 \ (\because a \neq 0) \)
\( 1+\lambda a=4+2 \mu \Rightarrow 2 \mu-a=-3\)
\( \Rightarrow a=3 \)
\( \therefore A(4,0,-1) \)
\( \therefore d^2=(\sqrt{16+1})^2=17\)

$$\begin{gathered} \frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}=\lambda \\ \text{Let} A \text{is the point on the line,} \\ A(2\lambda +1,-\lambda -2,2\lambda -3) \text{and} P(2,-10,1) \\ \because \overrightarrow{\mathrm{PA}}·\overrightarrow{\mathrm{n}}=0 \\ \Rightarrow (2\lambda -1)2+(-\lambda +8)(-1)+(2\lambda -4)2=0 \\ \Rightarrow 4\lambda -2+\lambda -8+4\lambda -8=0 \\ \Rightarrow 9\lambda -18=0\Rightarrow \lambda =2 \\ \therefore \mathrm{A}(5,-4,1) \\ \therefore \mathrm{AP}=\sqrt{3^2+6^2+0^2}=\sqrt{45}=3\sqrt{5} \end{gathered}$$

Shortest distance \(=\frac{\left|\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)\right|}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}\)

\(\overrightarrow{b_1} \times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5\end{array}\right|\)

\(=\hat{i}(-10+12)-\hat{j}(-5+6)+\hat{k}(4-4)\)

\(=2 \hat{i}-\hat{j}+0 \hat{k}\)

\(\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=\sqrt{2^2+(-1)^2+0^2}=\sqrt{5}\)

\(\overrightarrow{a_2}-\overrightarrow{a_1}=(-2-4) \hat{i}+(6-3) \hat{j}+(5-2) \hat{k}=-6 \hat{i}+3 \hat{j}+3 \hat{k}\)

\(\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)\)

\(=(-6 \hat{i}+3 \hat{j}+3 \hat{k})\cdot (2 \hat{i}-\hat{j}+0 \hat{k})=-15\)

Hence shortest distance

\(d=\frac{15}{\sqrt{5}}=3 \sqrt{5}\)

\(\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}=\lambda\)

\( x=2 \lambda-1, y=3 \lambda-1 \) & \(z=6 \lambda-3\)

\(\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}\)

\( x=2 \mu-1, y=3 \mu-1 \) & \(z=9\)

Let \(M\) is the point of intersection of both given lines

$\Rightarrow 2\lambda -1=2\mu -1,3\lambda -1=3\mu -1,6\lambda -3=9$

$\Rightarrow \lambda =2=\mu$

$\Rightarrow \mathit{\text{M }}\left(3, 5, 9\right)$

Now let point $P$ be $\left(2\text{ K}-1,3\text{ K}-1,6\text{ K}-3\right)$ on $\text{L}$ such that $\text{PM}=7$

$\Rightarrow \sqrt{{(2\text{K}-4)}^2+{(3\text{K}-6)}^2+{(6\text{K}-12)}^2}=7$

$\Rightarrow 49{\text{ K}}^2+196-196\text{K}=49$

$\Rightarrow {\text{K}}^2+4-4\text{K}=1$

$\Rightarrow {\text{K}}^2-4\text{K}+3=0$

$\Rightarrow \text{K}=1, 3$

So points are $P\left(1, 2, 3\right)$ or $P\left(5, 8, 15\right)$

So sum of all co-ordinates of \(P = 34\)

Given

$$\begin{gathered} L_1:\overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda (\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{r}}=(\lambda -1)\hat{\mathrm{i}}+2(\lambda +1)\hat{\mathrm{j}}+(\lambda +1)\hat{\mathrm{k}} \end{gathered}$$

Next,

$$\begin{gathered} L_2:\overrightarrow{\mathrm{r}}=(\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu (2\hat{\mathrm{i}}+7\hat{\mathrm{j}}+3\hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{r}}=2\mu \hat{\mathrm{i}}+(1+7\mu )\hat{\mathrm{j}}+(1+3\mu )\hat{\mathrm{k}} \end{gathered}$$

for point of intersection of $L_1 \text{and} L_2$

$$\begin{gathered} \lambda -1=2\mu \text{and} 2(\lambda +1)=1+7\mu \\ \text{On solving, }\lambda =3\text{ and }\mu =1 \\ \Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=3\hat{\mathrm{i}}+9\hat{\mathrm{j}}+4\hat{\mathrm{k}} \\ \text{hence, } \\ {\mathrm{L}}_3:\overrightarrow{\mathrm{r}}=2\hat{\mathrm{i}}+8\hat{\mathrm{j}}+4\hat{\mathrm{k}}+\alpha (3\hat{\mathrm{i}}+9\hat{\mathrm{j}}+4\hat{\mathrm{k}}) \\ \text{from options, we can see} {\mathrm{L}}_3 \mathrm{passes} \\ \mathrm{through} (8,26,12) \end{gathered}$$

Direction cosines of two lines satisfy the equation

$$\begin{gathered} 4\ell +m-n=0 \\ \Rightarrow n=4\ell +m ...\left(1\right) \\ 2mn+10n\ell +3\ell m=0 \\ \Rightarrow n\left(2m+10\ell \right)+3\ell m=0 ...\left(2\right) \end{gathered}$$

from eq. (1) and (2):

$$\begin{gathered} \Rightarrow (4\ell +m)(2m+10\ell )+3\ell m=0 \\ \Rightarrow 8\ell m+40{\ell }^2+2m^2+10\ell m+3\ell m=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow 40{\ell }^2+21\ell m+2m^2=0 \\ \Rightarrow (8\ell +m)(5\ell +2m)=0 \end{gathered}$$

Case-1: $8\ell +m=0\Rightarrow m=-8\ell$
So direction ratio of $L_1$ is $\ell ,-8\ell ,-4\ell$

Case-2: $5\ell +2m=0\Rightarrow m=\frac{-5}{2}\ell$
direction ratio of $L_2$ is $\ell ,\frac{-5\ell }{2},\frac{3\ell }{2}$

$$\begin{gathered} \cos \theta =\left|\frac{{\ell }^2+20{\ell }^2-6{\ell }^2}{\sqrt{{\ell }^2+64{\ell }^2+16{\ell }^2}\sqrt{{\ell }^2+\frac{25{\ell }^2}{4}+\frac{9{\ell }^2}{4}}}\right| \\ =\frac{15{\ell }^2}{\left(9\ell \right)\frac{\sqrt{38}\ell }{2}}=\frac{10}{3\sqrt{38}} \\ =\frac{10}{3\sqrt{38}} \end{gathered}$$

$\Rightarrow \overrightarrow{p}\times \overrightarrow{q}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|=-\widehat{i}+2\hat{j}-k$

$A\equiv \left(1,2,3\right)B\equiv \left(\lambda ,4,5\right)$

Shortest Distance

$=\left|\frac{\overrightarrow{AB}\cdot \left(\overrightarrow{p}\times \overrightarrow{q}\right)}{\left|\overrightarrow{p}\times \overrightarrow{q}\right|}\right|.\frac{1}{\sqrt{6}}$

$=\left|\frac{\left(\left(\lambda -1\right)\widehat{i}-2\widehat{j}+2\widehat{k}\right)\cdot \left(-\widehat{i}+2\widehat{j}-\widehat{k}\right)}{\sqrt{6}}\right|$

$\Rightarrow \left|-\lambda +1+4-2\right|=1$

$\Rightarrow \left|\lambda -3\right|=1$

$\Rightarrow \lambda =3\pm 1=4,2$

Radius of circle passing through points (0, 0), (4, 2) & (2, 4)

$=\frac{abc}{4\Delta }$

$=\frac{\sqrt{20}\times \sqrt{20}\times \sqrt{8}}{4\times \frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ 0& 4& 2\\ 0& 2& 4\end{array}\right|}$

$=\frac{20\times 2\sqrt{2}}{2\times 12}=\frac{5\sqrt{2}}{3}$

$$\begin{gathered} P_1:x-2y-2z+1=0 \\ {P}_2:2x-3y-6z+1=0 \\ \left|\frac{x-2y-2z+1}{\sqrt{1+4+4}}\right|=\left|\frac{2x-3y-6z+1}{\sqrt{2^2+3^2+6^2}}\right| \\ \frac{x-2y-2z+1}{3}=\pm \frac{2x-3y-6z+1}{7} \\ \text{ Since }{a}_1{a}_2+b_1{b}_2+c_1{c}_2=20>0 \\ \text{∴Negative sign will give acute bisector} \\ 7 x-14 y-14 z+7=-[6 x-9 y-18 z+3] \\ \Rightarrow 13x-23y-32z+10=0 \\ \left(-2,0,-\frac{1}{2}\right)\text{ satisfy it. } \end{gathered}$$

$$\begin{gathered} {\text{ L}}_1\text{: }\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}, L_2:\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1} \\ D\text{'}ratio of L_1\times L_2=\left|\begin{array}{ccc}\hat{ı}& \hat{ȷ}& \hat{k}\\ 1& -1& 2\\ -1& 2& 1\end{array}\right|=-5\hat{ı}-3\hat{ȷ}+\hat{k} \\ {L}_3:\frac{x-\alpha }{-5}=\frac{y-\beta }{-3}=\frac{z-\gamma }{1}=\mu \\ \left(5\mu +\alpha ,3\mu +\beta ,-\mu +\gamma \right) \\ P at L_1, \left(\lambda +1,-\lambda +2,2\lambda +1\right) \\ 5\mu +\alpha =\lambda +1 3\mu +\beta =-\lambda +2 -\mu +\gamma =2\lambda +1 \\ 5\mu -\lambda =1-\alpha 3\mu +\lambda =2-\beta \\ \mu =\frac{3-\alpha -\beta }{8} then \lambda =\frac{7+3\alpha -5\beta }{8} \\ -\left(\frac{3-\alpha -\beta }{8}\right)+\gamma =2\left(\frac{7+3\alpha -5\beta }{8}\right)+1 \\ -3+\alpha +\beta +8\gamma =14+6\alpha -10\beta +8 \\ \left|5\alpha -11\beta -8\gamma \right|=\left|-25\right|=25 \end{gathered}$$

G is centroid of \(\triangle PQR\)

$\therefore G\equiv \left(\frac{2\times 2-1}{3},\frac{2\times 1+4}{3},\frac{2\times 2+2}{3}\right)$

$G\equiv (1,2,2)$

Let \(x\) is the point of intersection of lines

$l_1:\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$

$k_2:\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1},\text{ then }x\equiv (2,-6,0)$

$\therefore$distance between $G$ and $x=\sqrt{69}$

G is centroid of \(\triangle PQR\)

$\therefore G\equiv \left(\frac{2\times 2-1}{3},\frac{2\times 1+4}{3},\frac{2\times 2+2}{3}\right)$

$G\equiv (1,2,2)$

Let \(x\) is the point of intersection of lines

$l_1:\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$

$k_2:\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1},\text{ then }x\equiv (2,-6,0)$

$\therefore$distance between $G$ and $x=\sqrt{69}$

Given \(P(2,-10,1)\)

$\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}=\lambda$

Let $A$ is the point on the line,

$A\left(2\lambda +1,-\lambda -2,2\lambda -3\right)$

$$\begin{gathered} \because \overrightarrow{\mathrm{PA}}·\overrightarrow{\mathrm{n}}=0 \\ \Rightarrow \left(2\lambda -1\right)2+\left(-\lambda +8\right)\left(-1\right)+\left(2\lambda -4\right)2=0 \\ \Rightarrow 4\lambda -2+\lambda -8+4\lambda -8=0 \\ \Rightarrow 9\lambda -18=0\Rightarrow \lambda =2 \\ \therefore \mathrm{A}\left(5,-4,1\right) \\ \therefore \mathrm{AP}=\sqrt{3^2+6^2+0^2}=\sqrt{45}=3\sqrt{5} \end{gathered}$$

Given point P(1,0,7).

Given line:

$\frac{\left(x-1\right)}{1}=\frac{y}{\left(-3\right)}=\frac{\left(z-8\right)}{1}$

A point on line is A(1,0,8).

Direction ratios of line = (1,−3,1).

Find foot of perpendicular H from P to line.

Using reflection formula:

$P^{\text{'}}=2H-P$

After applying angle conditions given in question, required point obtained is the option corresponding to (B).

Given point P(1,0,7).

Given line:

$\frac{\left(x-1\right)}{1}=\frac{y}{\left(-3\right)}=\frac{\left(z-8\right)}{1}$

A point on line is A(1,0,8).

Direction ratios of line = (1,−3,1).

Find foot of perpendicular H from P to line.

Using reflection formula:

$P^{\text{'}}=2H-P$

After applying angle conditions given in question, required point obtained is the option corresponding to (B).

\(\begin{aligned}
& A \equiv(1,2,1), C(-2,2,-1) \\
& \text { where } \hat{\mathrm{n}}=\left|\begin{array}{lll}
\hat{\mathrm{i}} & \hat{j} & \hat{k} \\
2 & 3 & 4 \\
7 & 8 & 2
\end{array}\right|=-26 \hat{\mathrm{i}}+24 \hat{\mathrm{j}}-5 \hat{\mathrm{k}} \\
& \\
& =\left|\frac{(3 \hat{\mathrm{i}}+2 \hat{k}) \cdot(26 \hat{\mathrm{i}}-24 \hat{\mathrm{j}}+5 \hat{k})}{\sqrt{1277}}\right| \\
& =\frac{78+10}{\sqrt{1277}}=\frac{88}{\sqrt{1277}}
\end{aligned}\)

$$\begin{gathered} \text{ Since L is perpendicular to given two lines} \\ So \text{Vector parallel to 'L' } \\ =\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 2& 1& -2\\ 1& 3& 4\end{array}\right|=10\hat{\mathrm{i}}-10\hat{\mathrm{j}}+5\hat{\mathrm{k}} \end{gathered}$$

$$\begin{gathered} =5(2\hat{i}-2\hat{j}+\hat{k}) \\ \text{ Equation of 'L' } \\ \frac{x-2}{2}=\frac{y+1}{-2}=\frac{z-3}{1}=\lambda \left(\text{ say }\right) \\ \text{ Let }\mathrm{Q}(2\lambda +2,-2\lambda -1,\lambda +3) \\ \text{Since it intersects with} yz \text{plane} \\ \Rightarrow 2\lambda +2=0\Rightarrow \lambda =-1 \\ \Rightarrow Q(0,1,2) \\ d(P, Q)=3 \end{gathered}$$

\(\begin{aligned}
& A \equiv(1,2,1), C(-2,2,-1) \\
& \text { where } \hat{\mathrm{n}}=\left|\begin{array}{lll}
\hat{\mathrm{i}} & \hat{j} & \hat{k} \\
2 & 3 & 4 \\
7 & 8 & 2
\end{array}\right|=-26 \hat{\mathrm{i}}+24 \hat{\mathrm{j}}-5 \hat{\mathrm{k}} \\
& \\
& =\left|\frac{(3 \hat{\mathrm{i}}+2 \hat{k}) \cdot(26 \hat{\mathrm{i}}-24 \hat{\mathrm{j}}+5 \hat{k})}{\sqrt{1277}}\right| \\
& =\frac{78+10}{\sqrt{1277}}=\frac{88}{\sqrt{1277}}
\end{aligned}\)

\(
\frac{x+1}{2}=\frac{y-1}{-1}=\frac{z+3}{2}=\lambda
\)

let Point Q\((2 \lambda-1,-\lambda+1,2 \lambda-3)\)
Now
\(P Q\) < \((2 \lambda+1,-\lambda+1,2 \lambda-5)\)>

\(
\begin{gathered}
P Q \perp\langle 2,-1,2\rangle \\
2(2 \lambda+1)\langle-1(-\lambda+1)+2(2 \lambda-5) \\
4 \lambda+2+\lambda-1+4 \lambda-10=0 \\
9 \lambda=9 \\
\lambda=1
\end{gathered}
\)

Q \((1,0,-1)\)

\(
\begin{aligned}
& P Q=\sqrt{(1+2)^2+(-1-2)^2} \\
& \sqrt{9+9}=3 \sqrt{2}
\end{aligned}
\)

$$\begin{gathered} \overrightarrow{{\mathrm{a}}_1}=\left(2,1,-3\right) \\ \overrightarrow{{\mathrm{a}}_2}=\left(-1,-3,-5\right) \\ \overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 1& 2& -3\\ 2& 4& -5\end{array}\right| \\ =2\hat{i}-\hat{j} \\ \overrightarrow{{\mathrm{a}}_1}-\overrightarrow{{\mathrm{a}}_2}=-3\hat{\mathrm{i}}-4\hat{\mathrm{j}}-2\hat{\mathrm{k}} \\ \mathrm{S}. \mathrm{D}. =\frac{\left|\right(\overrightarrow{{\mathrm{a}}_1}-\overrightarrow{{\mathrm{a}}_2})·(\overrightarrow{{\mathrm{b}}_1}\times \overrightarrow{{\mathrm{b}}_2}\left)\right|}{|\overrightarrow{{\mathrm{b}}_1}\times \overrightarrow{{\mathrm{b}}_2}|} \\ =\frac{2}{\sqrt{5}} \\ {\left(\mathrm{S}.\mathrm{D}.\right)}^2=\frac{4}{5} \\ \mathrm{m}=4,\mathrm{n}=5\Rightarrow \mathrm{m}+\mathrm{n}=9 \end{gathered}$$

• Point $P$: $(0, 2, 3)$
• Base $QR$: The length of the base of the triangle is given as $QR = 5$.
• Line $L$: The vertices $Q$ and $R$ lie on the line: $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$ From this equation, we can determine:
  • A point on the line, let's call it $A = (-3, 1, -4)$.
  • The direction vector of the line, $\vec{d} = (5, 2, 3)$.

The height of the triangle $PQR$ is the perpendicular distance from point $P$ to the line $L$.

Let $\vec{v}$ be the vector from point $A$ to point $P$:

$\vec{v}=P-A=(0-(-3),2-1,3-(-4\left)\right)$$=(3,1,7)$

$h = \frac{|\vec{v} \times \vec{d}|}{|\vec{d}|}$

$\vec{v}\times \vec{d}=\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 3& 1& 7\\ 5& 2& 3\end{array}\mid$$=\mathbf{i}(3-14)-\mathbf{j}(9-35)+\mathbf{k}(6-5)$$=-11\mathbf{i}+26\mathbf{j}+1\mathbf{k}$

The magnitude of the cross product is:

$\mathrm{\mid }\vec{v}\times \vec{d}\mathrm{\mid }=\sqrt{(-11{)}^2+{26}^2+1^2}$$=\sqrt{121+676+1}=\sqrt{798}$

$|\vec{d}| = \sqrt{5^2 + 2^2 + 3^2} = \sqrt{25 + 4 + 9} = \sqrt{38}$

$h = \sqrt{\frac{798}{38}} = \sqrt{21}$

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ $\text{Area} = \frac{1}{2} \times 5 \times \sqrt{21} = \frac{5\sqrt{21}}{2}$

$\frac{m}{n} = \frac{5\sqrt{21}}{2}$

$2m-5\sqrt{21}n=0$

Let point on the line \(\frac{x-1}{2}=\frac{y+1}{-3}=z\)

is $P\left(2\lambda +1,-3\lambda -1,\lambda \right)$

Now distance of \(P\) from the point \((1,-1,0)\) is \(4 \sqrt{14}\)

Then $4{\lambda }^2+9{\lambda }^2+{\lambda }^2=16\cdot 14$

$\Rightarrow \lambda =\pm 4$

\(\mathrm{P}(9,-13,4)\) or \(\mathrm{P}(-7,11,-4)\)(nearer to origin)

\(\therefore \alpha=-7, \beta=11, \gamma=-4\)

Now

\(L_1: \frac{x+7}{1}=\frac{y-11}{2}=\frac{z+4}{3}\)

\(L_2: \frac{x+5}{2}=\frac{y-10}{1}=\frac{z-3}{1}\)

$\therefore$ Shortest distance $=\left|\frac{\left|\begin{array}{ccc}2& -1& 7\\ 1& 2& 3\\ 2& 1& 1\end{array}\right|}{\left|\begin{array}{lll}\widehat{i}& \widehat{j}& \widehat{k}\\ 1& 2& 3\\ 2& 1& 1\end{array}\right|}\right|$

$=\frac{28}{\sqrt{1+25+9}}=\frac{4\sqrt{7}}{\sqrt{5}}$

$\Rightarrow \overrightarrow{p}\times \overrightarrow{q}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|=-\widehat{i}+2\hat{j}-k$

$A\equiv \left(1,2,3\right)B\equiv \left(\lambda ,4,5\right)$

Shortest Distance$=\left|\frac{\overrightarrow{AB}\cdot \left(\overrightarrow{p}\times \overrightarrow{q}\right)}{\left|\overrightarrow{p}\times \overrightarrow{q}\right|}\right|$

$\left|\frac{\left(\left(\lambda -1\right)\widehat{i}-2\widehat{j}+2\widehat{k}\right)\cdot \left(-\widehat{i}+2\widehat{j}-\widehat{k}\right)}{\sqrt{6}}\right|\text{=}\frac{1}{\sqrt{6}}$

$\Rightarrow \left|-\lambda +1+4-2\right|=1$

$\Rightarrow \left|\lambda -3\right|=1$

$\Rightarrow \lambda =3\pm 1=4,2$

Radius of circle passing through points \((0, 0), (4, 2)\) and \((2, 4)\)

using circumradius: $R=\frac{abc}{4\Delta }$

$=\frac{\sqrt{20}\times \sqrt{20}\times \sqrt{8}}{4\times \frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ 0& 4& 2\\ 0& 2& 4\end{array}\right|}$

$=\frac{20\times 2\sqrt{2}}{2\times 12}=\frac{5\sqrt{2}}{3}$

$$\begin{gathered} {\text{ L}}_1\text{: }\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}, L_2:\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1} \\ \text{DR of} L_1\times L_2=\left|\begin{array}{ccc}\hat{ı}& \hat{ȷ}& \hat{k}\\ 1& -1& 2\\ -1& 2& 1\end{array}\right|=-5\hat{ı}-3\hat{ȷ}+\hat{k} \\ {\text{L}}_3:\frac{x-\alpha }{-5}=\frac{y-\beta }{-3}=\frac{z-\gamma }{1}=\mu \\ \left(5\mu +\alpha ,3\mu +\beta ,-\mu +\gamma \right) \\ P \text{at} L_1, \left(\lambda +1,-\lambda +2,2\lambda +1\right) \\ 5\mu +\alpha =\lambda +1 3\mu +\beta =-\lambda +2 -\mu +\gamma =2\lambda +1 \\ 5\mu -\lambda =1-\alpha 3\mu +\lambda =2-\beta \\ \mu =\frac{3-\alpha -\beta }{8}\text{ then} \lambda =\frac{7+3\alpha -5\beta }{8} \\ -\left(\frac{3-\alpha -\beta }{8}\right)+\gamma =2\left(\frac{7+3\alpha -5\beta }{8}\right)+1 \\ -3+\alpha +\beta +8\gamma =14+6\alpha -10\beta +8 \\ \left|5\alpha -11\beta -8\gamma \right|=\left|-25\right|=25 \end{gathered}$$

Direction cosines of two lines satisfy the equation

$$\begin{gathered} 4\ell +m-n=0 \\ \Rightarrow n=4\ell +m ...\left(1\right) \\ 2mn+10n\ell +3\ell m=0 \\ \Rightarrow n\left(2m+10\ell \right)+3\ell m=0 ...\left(2\right) \end{gathered}$$

from eq. (1) and (2):

$$\begin{gathered} \Rightarrow (4\ell +m)(2m+10\ell )+3\ell m=0 \\ \Rightarrow 8\ell m+40{\ell }^2+2m^2+10\ell m+3\ell m=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow 40{\ell }^2+21\ell m+2m^2=0 \\ \Rightarrow (8\ell +m)(5\ell +2m)=0 \end{gathered}$$

Case-1: $8\ell +m=0\Rightarrow m=-8\ell$
So direction ratio of $L_1$ is $\ell ,-8\ell ,-4\ell$

Case-2: $5\ell +2m=0\Rightarrow m=\frac{-5}{2}\ell$
direction ratio of $L_2$ is $\ell ,\frac{-5\ell }{2},\frac{3\ell }{2}$

$$\begin{gathered} \cos \theta =\left|\frac{{\ell }^2+20{\ell }^2-6{\ell }^2}{\sqrt{{\ell }^2+64{\ell }^2+16{\ell }^2}\sqrt{{\ell }^2+\frac{25{\ell }^2}{4}+\frac{9{\ell }^2}{4}}}\right| \\ =\frac{15{\ell }^2}{\left(9\ell \right)\frac{\sqrt{38}\ell }{2}}=\frac{10}{3\sqrt{38}} \\ =\frac{10}{3\sqrt{38}} \end{gathered}$$

Let \(P(\lambda, 2,3)\)

\(\frac{x-4}{1}=\frac{y-9}{2}=\frac{z-5}{1}\)

\(\mathrm{Q}(1, \mu, 2)\) satisfies the line \(L\)

\(\therefore \frac{1-4}{1}=\frac{\mu-9}{2}=\frac{2-5}{1}\)

\(\Rightarrow-3=\frac{\mu-9}{2}=-3 \Rightarrow \mu=3\)

D.R. of \(\mathrm{PQ}: \lambda-1,2-\mu, 1\)

Since \(\mathrm{PQ} \perp \mathrm{L} \Rightarrow(\lambda-1) \cdot 1+2(2-\mu)+1\cdot 1=0\)

\(\Rightarrow \lambda-2 \mu+4=0 \)

\(\therefore \lambda=2\)

Now lines : \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}\)

\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z+5}{6}\)

i.e. \(\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+t_1(2 \hat{i}+3 \hat{j}+6 \hat{k})\)

\(\vec{r}=(2 \hat{i}+3 \hat{j}-5 \hat{k})+t_2(2 \hat{i}+3 \hat{j}+6 \hat{k})\)

Now distance

\(=\frac{|((\hat{i}+2 \hat{j}-4 \hat{k})-(2 \hat{i}+3 \hat{j}-5 \hat{k})) \times(2 \hat{i}+3 \hat{j}+6 \hat{k})|}{|2 \hat{i}+3 \hat{j}+6 \hat{k}|}\)

\(=\frac{|-9 \hat{i}+8 \hat{j}-\hat{k}|}{7}=\frac{\sqrt{146}}{7}\)

Direction ratios of \(L_1\) are

\(\vec{d}_1=(3,5,7)\)

Direction ratios of \(L_2\) are

\(\vec{d}_2=(1,4,7)\)

Since line \(L\) is perpendicular to both, its direction vector will be parallel to

\(\vec{d}=\vec{d}_1 \times \vec{d}_2\)

Now, \(\vec{d}_1 \times \vec{d}_2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 7 \\ 1 & 4 & 7\end{array}\right|\)

\(=7 \hat{i}-14 \hat{j}+7 \hat{k}\)

So direction ratios of line \(L\) can be taken as \((1,-2,1)\)

Direction ratios of \(L_3\) are

\((2,1,2)\)

Let \(\theta\) be the acute angle between \(L\) and \(L_3\)

\(\tan \theta=\frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|}\)

Where \(\vec{a}=(1,-2,1), \vec{b}=(2,1,2)\)

\(\vec{a} \cdot \vec{b}=1 \cdot 2+(-2) \cdot 1+1 \cdot 2=2\)

\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 2 & 1 & 2\end{array}\right|\)

\(=-5 \hat{i}+0 \hat{j}+5 \hat{k}\)

\(|\vec{a} \times \vec{b}|=\sqrt{(-5)^2+0^2+5^2}=\sqrt{50}=5 \sqrt{2}\)

Hence, \(\tan \theta=\frac{5 \sqrt{2}}{2}\)

Given

shortest distance between the lines \(\frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha}\) and \(\frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2 \alpha}\) is \(\sqrt{2}\)

${\sqrt{2}=\left|\frac{\left|\begin{array}{ccc}-1& 1& 3\\ \alpha & -1& -\alpha \\ \alpha & 2& 2\alpha \end{array}\right|}{\left|\begin{array}{ccc}\widehat{i}& j& k\\ \alpha & -1& -\alpha \\ \alpha & 2& 2\alpha \end{array}\right|}\right|}_{}$

$\sqrt{2}=\frac{\left|-1(-2\alpha +2\alpha )-1\left(2{\alpha }^2+{\alpha }^2\right)+3(2\alpha +\alpha )\right|}{\left|\widehat{i}(-2\alpha +2\alpha )-\widehat{j}\left(2{\alpha }^2+{\alpha }^2\right)+\widehat{k}(2\alpha +\alpha )\right|}$

$\sqrt{2}=\frac{\left|-3{\alpha }^2+9\alpha \right|}{\sqrt{9{\alpha }^4+9{\alpha }^2}}$

$\sqrt{2}=\frac{\left|-\alpha +3\right|}{\sqrt{{\alpha }^2+1}}$

$\Rightarrow 2{\alpha }^2+2={\alpha }^2+9-6\alpha$

${\alpha }^2+6\alpha -7=0$

$(\alpha +7)(\alpha -1)=0$

$\alpha =-7,1$

sum $=–7+1=–6$

$$\begin{gathered} \text{Let the line parallel to }\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3} \\ \text{passing through }(1,4,0)\text{ is } \\ \frac{x-1}{1}=\frac{y-4}{2}=\frac{z-0}{3} \\ \text{so their point of intersection is } \\ \text{calculated by equating their general point} \\ (\lambda +1,2\lambda +4,3\lambda )=(2t+2,3t+6,4t+3) \\ \lambda =2t+1 \\ 2\lambda +4=3t+6\Rightarrow t=0 \\ \text{ so Point of intersection is }(2,6,3) \\ \text{ so distance }=\sqrt{(2-1{)}^2+(6-4{)}^2+(3-0{)}^2}=\sqrt{14} \end{gathered}$$

\begin{aligned}
& \overrightarrow{a_1}=2 \hat{\jmath}+\hat{\jmath}-3 \hat{k} \\
& \overrightarrow{a_2}=-\hat{l}-3 \hat{\jmath}-5 \hat{k} \\
& \overrightarrow{a_2}-\overrightarrow{a_1}=-3 \hat{l}-4 \hat{\jmath}-2 \hat{k} \\
& \overrightarrow{b_1}=\hat{l}+2 \hat{\jmath}-3 \hat{k} \\
& \quad \overrightarrow{b_1} \times \overrightarrow{b_2}=\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{k}+4 \hat{\jmath}-5 \hat{k} \\
1 & 2 & -3 \\
2 & 4 & -5
\end{array}\right|=2 \hat{l}-\hat{\jmath}
\end{aligned}

\(\begin{aligned}
& \left.S.D=\left|\frac{(-3 \hat{i}-4 \hat{\jmath}-2 \hat{k}) \cdot(2 \hat{l}-\hat{\jmath}) \mid}{|2 \hat{\imath}-\hat{j}|}\right| \frac{\mid-6+4}{\sqrt{5}} \right\rvert\,=\frac{2}{\sqrt{5}} \\
& \text { square of }SD=\frac{4}{5}=\frac{m}{n} \\
& m=4 \quad n=5 \\
& m+n=4+5=9
\end{aligned}\)

Let \(\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}=t\)

So any point on the line is

\((x, y, z)=(-3+8 t, 4+2 t,-1+2 t)\)

Its distance from \(R(1,2,3)\) is \(6\),

so \((-3+8 t-1)^2+(4+2 t-2)^2+(-1+2 t-3)^2=36 \)
\((8 t-4)^2+(2 t+2)^2+(2 t-4)^2=36 \)
\(72 t^2-72 t+36=36 \)
\(72 t(t-1)=0\)
So, \(t=0,1\)

Thus, \(P=(-3,4,-1), Q=(5,6,1)\)

Centroid of \(\triangle P Q R\) is:

\((\alpha, \beta, \gamma)=\left(\frac{-3+5+1}{3}, \frac{4+6+2}{3}, \frac{-1+1+3}{3}\right)\)
\((\alpha, \beta, \gamma)=(1,4,1) \)
\(\alpha+\beta+\gamma=1+4+1=6\)

$L_1:(7\widehat{i}+6\widehat{j}+2\widehat{k})-\lambda (-3\widehat{i}+2\widehat{j}+4\widehat{k})$

$L_1:(5\widehat{i}+3\widehat{j}+4\widehat{k})+\lambda (2\widehat{i}+\widehat{j}+3\widehat{k})$

Distance between skew lines

$=\frac{(2\widehat{i}+3\widehat{j}-2\widehat{k})\cdot (2\widehat{i}+17\widehat{j}-7\widehat{k})}{\sqrt{342}}$

$=\frac{69}{\sqrt{342}}=\frac{69}{3\sqrt{38}}=\frac{23}{\sqrt{38}}$

\(\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}=\lambda\)

\( x=2 \lambda-1, y=3 \lambda-1 \) & \(z=6 \lambda-3\)

\(\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}\)

\( x=2 \mu-1, y=3 \mu-1 \) & \(z=9\)

Let \(M\) is the point of intersection of both given lines

$\Rightarrow 2\lambda -1=2\mu -1,3\lambda -1=3\mu -1,6\lambda -3=9$

$\Rightarrow \lambda =2=\mu$

$\Rightarrow \mathit{\text{M }}\left(3, 5, 9\right)$

Now let point $P$ be $\left(2\text{ K}-1,3\text{ K}-1,6\text{ K}-3\right)$ on $\text{L}$ such that $\text{PM}=7$

$\Rightarrow \sqrt{{(2\text{K}-4)}^2+{(3\text{K}-6)}^2+{(6\text{K}-12)}^2}=7$

$\Rightarrow 49{\text{ K}}^2+196-196\text{K}=49$

$\Rightarrow {\text{K}}^2+4-4\text{K}=1$

$\Rightarrow {\text{K}}^2-4\text{K}+3=0$

$\Rightarrow \text{K}=1, 3$

So points are $P\left(1, 2, 3\right)$ or $P\left(5, 8, 15\right)$

So sum of all co-ordinates of \(P = 34\)

Given point \(A(-2,-8,6)\)

Direction ratios of the line along which distance is measured are \((1,-1,2)\)

Hence the line through \(A\) parallel to these DRs is

\(\frac{x+2}{1}=\frac{y+8}{-1}=\frac{z-6}{2}=\lambda\)

So a general point on this line is \(B(\lambda-2,-\lambda-8,2 \lambda+6)\)

The given line is \(\frac{x-1}{1}=\frac{y-1}{2}=\frac{z}{-1}=\mu\)

Hence a general point on it is \(B(\mu+1,2 \mu+1,-\mu)\)

Since both represent the same point \(B\),

On comparing we get point \(B\) is \((-3,-7,4)\)

\( \mathrm{AB}=\sqrt{(-1)^2+(1)^2+(-2)^2}=\sqrt{6} \)
\( \mathrm{AB}^2=6\)