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:

The line $$\frac{x-2}{3} = \frac{y+1}{-2} = \frac{z+3}{-1}$$ lies on the plane $$px - qy + z = 5$$. The point $$(2, -1, -3)$$ on this line satisfies the plane equation, giving $$2p - (-1)q + (-3) = 5$$, or $$2p + q = 8$$. Since the direction vector $$(3, -2, -1)$$ of the line must be perpendicular to the normal vector $$(p, -q, 1)$$ of the plane, their dot product yields $$3p + 2q - 1 = 0$$. Solving the system $$2p + q = 8$$ and $$3p + 2q - 1 = 0$$ by substituting $$q = 8 - 2p$$ gives $$p = 15$$ and $$q = -22$$, so the required plane is $$15x + 22y + z = 5$$.

The shortest distance from the origin to this plane is $$d = \frac{|15(0) + 22(0) + 1(0) - 5|}{\sqrt{15^2 + 22^2 + 1^2}} = \frac{5}{\sqrt{710}} = \sqrt{\frac{5}{142}}$$. Therefore, the answer is Option B: $$\sqrt{\frac{5}{142}}$$.