${\mathrm{L}}_{1}=\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2},{L}_{2}:\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}$ Let the line ${\mathrm{L}}_{3}$ passes through the point $(\alpha ,\beta ,\gamma )$ perpendicular to ${L}_{1}&{L}_{2}$ and ${L}_{3}$ intersect line ${L}_{1}$ then $|5\alpha -11\beta -8\gamma |$.

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YESPYQ · Accepted Answer

Correct answer: a) 25. ${\text{ L}}_{1}\text{: }\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2},{L}_{2}:\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}\ D'ratioof{L}_{1}\times {L}_{2}=\left|\begin{matrix}\overset{^}{ı} & \overset{^}{ȷ} & \overset{^}{k} \ 1 & -1 & 2 \ -1 & 2 & 1\end{matrix}\right|=-5\overset{^}{ı}-3\overset{^}{ȷ}+\overset{^}{k}\ {L}_{3}:\frac{x-\alpha }{-5}=\frac{y-\beta }{-3}=\frac{z-\gamma }{1}=\mu \ \left(5\mu +\alpha ,3\mu +\beta ,-\mu +\gamma \right)\ Pat{L}_{1},\left(\lambda +1,-\lambda +2,2\lambda +1\right)\ 5\mu +\alpha =\lambda +13\mu +\beta =-\lambda +2-\mu +\gamma =2\lambda +1\ 5\mu -\lambda =1-\alpha 3\mu +\lambda =2-\beta \ \mu =\frac{3-\alpha -\beta }{8}then\lambda =\frac{7+3\alpha -5\beta }{8}\ -\left(\frac{3-\alpha -\beta }{8}\right)+\gamma =2\left(\frac{7+3\alpha -5\beta }{8}\right)+1\ -3+\alpha +\beta +8\gamma =14+6\alpha -10\beta +8\ \left|5\alpha -11\beta -8\gamma \right|=\left|-25\right|=25$

\({\mathrm{L}}_{1}=\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2},{L}_{2}:\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}\) Let the…

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