Let $\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1}$ lie on the plane $px-qy+z=5$, for some $p, q\in R$. The shortest distance of  the plane from the origin is:

Given, line $\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1}$

So, the point $\left(2, -1, -3\right)$ will satisfy the given plane $px-qy+z=5$

So, $2p+q-3=5$ or $2p+q=8$ ... (i)

Also given line is perpendicular to normal plane so 

$3p+2q-1=0$ ... (ii)

$\Rightarrow p=15, q=-22$

So,equation of plane $15x-22y+z-5=0$

Now its distance from origin $=\frac{6}{\sqrt{710}}=\sqrt{\frac{5}{142}}$