Let $$P$$ be the point of intersection of the lines $$\frac{x-2}{1} = \frac{y-4}{5} = \frac{z-2}{1}$$ and $$\frac{x-3}{2} = \frac{y-2}{3} = \frac{z-3}{2}$$. Then, the shortest distance of $$P$$ from the line $$4x = 2y = z$$ is

Line 1: (2+t,4+5t,2+t). Line 2: (3+2s,2+3s,3+2s). Intersection: 2+t=3+2s,4+5t=2+3s,2+t=3+2s. From 1st and 3rd: same. 2+t=3+2s → t=1+2s. 4+5(1+2s)=2+3s → 9+10s=2+3s → 7s=-7 → s=-1.

t=-1. P=(1,-1,1). Line 4x=2y=z: direction (1,2,4), point (0,0,0).

$$\vec{AP}=(1,-1,1)$$. Projection on (1,2,4): $$(1-2+4)/\sqrt{21}=3/\sqrt{21}$$.

Distance=$$\sqrt{3-9/21}=\sqrt{3-3/7}=\sqrt{18/7}=3\sqrt{2/7}=3\sqrt{14}/7$$.

The answer is Option (2): $$\frac{3\sqrt{14}}{7}$$.