The distance of the point $(-1,2,-2)$ from the line of intersection of the planes $2x+3y+2z=0$ and $x-2y+z=0$ is

The correct option is B $\frac{\sqrt{34}}{2}$


DRs of line of intersection
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 2\\ 1& -2& 1\end{array}\right|=7\hat{i}-0\hat{j}-7\hat{k}$
DRs $=(1,0,-1)$
Both the given planes pass through $(0,0,0)$, so line of intersection also passes through $(0,0,0)$.
Equation of line of intersection
$\frac{x}{1}=\frac{y}{0}=\frac{z}{-1}=r$
$M(r,0,-r)$
Direction ratios of $PM=(r+1,-2,-r+2)$
$\because PM$ is perpendicular to the line of intersection
$\therefore 1(r+1)+0(-2)-1(-r+2)=0$
$\Rightarrow 2r-1=0$
$\Rightarrow r=\frac{1}{2}$
$M(\frac{1}{2},0,\frac{-1}{2})$
$PM=\sqrt{\frac{9}{4}+4+\frac{9}{4}}=\frac{\sqrt{34}}{2}$