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

(2, -1, -3) satisfy the given plane.
So 2p + q = 8 .... (i)
Also given line is perpendicular to normal plane so
3p + 2q - 1 = 0 .... (ii)
$\Rightarrow \mathrm{p}=15,\mathrm{q}=-22$
Eq. of plane 15x - 22y + z - 5 = 0
its distance from origin $\frac{5}{\sqrt{{15}^2+{22}^2+1}}=\frac{5}{\sqrt{710}}=\sqrt{\frac{5}{142}}$