Let (x-2)/3=(y+1)/-2=(z+3)/-1 lie on the plane p x-q y+z=5, for some p, q ∈ R. The shortest distance of the plane from the origin is:

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