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 line $l_1$ passes through the point $(2, 6, 2)$ and is perpendicular to the line $2x+y-2z=10$ then,

$l_1: \frac{x-2}{2}=\frac{y-6}{1}=\frac{z-2}{-2}$ 

Let the shortest distance between $l_1$ and the line $\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$ is S.D.

S.D. $=\left|\frac{\overrightarrow{AB}·\overrightarrow{MN}}{\left|\overrightarrow{MN}\right|}\right|$

Here, $\overrightarrow{AB}=3\hat{i}+10\hat{j}+2\hat{k}$

$\overrightarrow{MN}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -2\\ 2& -3& 2\end{array}\right|=-4\hat{i}-8\hat{j}-8\hat{k}$

$\Rightarrow \left|\overrightarrow{MN}\right|=\sqrt{16+64+64}=12$

$\overrightarrow{AB}·\overrightarrow{MN}=-12-80-16=-108$

So, shortest distance (S.D.) 

$=\left|\frac{-108}{12}\right|=\left|-9\right|=9$ units