Let the values of $\lambda$ for which the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-\lambda }{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is $\frac{1}{\sqrt{6}}$ be ${\lambda }_{1}$ and ${\lambda }_{2}$. Then the radius of the circle passing through the points $(0,0),\left({\lambda }_{1},{\lambda }_{2}\right)$ and $\left({\lambda }_{2},{\lambda }_{1}\right)$ is [JEE Main 2025, 8 Apr (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: a) $\frac{5\sqrt{2}}{3}$. $\Rightarrow \text{ }\vec{p}\text{ }\times \text{ }\vec{q}=\text{ }\left|\begin{matrix}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \ 2 & 3 & 4 \ 3 & 4 & 5\end{matrix}\right|\text{ }=−\overset{^}{i}+2\overset{^}{j}-k$ $A\text{ }\equiv \text{ }\left(1,\text{ }2,\text{ }3\right)\text{ }B\text{ }\equiv \text{ }\left(\lambda ,\text{ }4,\text{ }5\right)$ Shortest Distance $=\text{ }\left|\frac{\vec{AB}\text{ }⋅\text{ }\left(\vec{p}\text{ }\times \vec{q}\right)}{\left|\vec{p}\text{ }\times \text{ }\vec{q}\right|}\right|\text{ }.\frac{1}{\sqrt{6}}$ $=\text{ }\left|\frac{\left(\left(\lambda −1\right)\text{ }\overset{^}{i}\text{ }−\text{ }2\overset{^}{j}\text{ }+\text{ }2\overset{^}{k}\right)\text{ }⋅\text{ }\left(−\overset{^}{i}\text{ }+\text{ }2\overset{^}{j}\text{ }−\text{ }\overset{^}{k}\right)}{\sqrt{6}}\right|$ $\Rightarrow \text{ }\left|−\text{ }\lambda +1+4−2\right|=1$ $\Rightarrow \text{ }\left|\text{ }\lambda −3\right|=1$ $\Rightarrow \text{ }\lambda =3\text{ }\pm \text{ }1=\text{ }4\text{ },\text{ }2\text{ }$ Radius of

Let the values of \(\lambda\) for which the shortest distance between the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}…

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