Thank you for reporting, we will resolve it shortly
Q.
The line $l_1$ passes through the point $(2,6,2)$ and is perpendicular to the plane $2 x+y-2 z=10$. Then the shortest distance between the line $l_1$ and the line $\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$ is :
Line $\ell$, is given by
$L _1: \frac{ x -2}{2}=\frac{ y -6}{1}=\frac{ z -2}{-2}$
Given,
$L _2: \frac{ x +1}{2}=\frac{ y +4}{-3}=\frac{ z }{2}$
Shortest distance $=\left|\frac{\overrightarrow{ AB } \cdot \overrightarrow{ MN }}{ MN }\right|$
$ \overrightarrow{ AB }=3 \hat{ i }+10 \hat{ j }+2 \hat{ k } $
$\overrightarrow{ MN }= \begin{vmatrix} \hat{ i } & \hat{ j } & \hat{ k } \\2 & 1 & -2 \\2 & -3 & 2\end{vmatrix}=-4 \hat{ i }-8 \hat{ j }-8 \hat{ k } $
$MN =\sqrt{16+64+64}=12$
$\therefore$ Shortest distance $=\left|\frac{-12-80-16}{12}\right|=9$
$\therefore$ Option (4) is correct.
