Consider the lines $$L_1$$ and $$L_2$$ given by
$$L_1: \frac{x-1}{2} = \frac{y-3}{1} = \frac{z-2}{2}$$
$$L_2: \frac{x-2}{1} = \frac{y-2}{2} = \frac{z-3}{3}$$
A line $$L_3$$ having direction ratios $$1, -1, -2$$, intersects $$L_1$$ and $$L_2$$ at the points $$P$$ and $$Q$$ respectively. Then the length of line segment $$PQ$$ is

$$L_1: \frac{x-1}{2} = \frac{y-3}{1} = \frac{z-2}{2}$$, $$L_2: \frac{x-2}{1} = \frac{y-2}{2} = \frac{z-3}{3}$$. Line $$L_3$$ has direction $$(1,-1,-2)$$ and intersects both.

We parameterize a point on $$L_1$$ as $$P = (1+2s, 3+s, 2+2s)$$ and a point on $$L_2$$ as $$Q = (2+t, 2+2t, 3+3t)$$. For $$\vec{PQ}$$ to be parallel to $$(1,-1,-2)$$, the ratios of its components must satisfy $$ \frac{(2+t)-(1+2s)}{1} = \frac{(2+2t)-(3+s)}{-1} = \frac{(3+3t)-(2+2s)}{-2}. $$ Equating the first two expressions, $$ 1+t-2s = -(-1+2t-s) = 1-2t+s, $$ which simplifies to $$t-2s = -2t+s$$ and hence $$3t = 3s \implies t = s.$$ Equating the first and third expressions, $$ \frac{1+t-2s}{1} = \frac{1+3t-2s}{-2} $$ leads to $$ -2(1+t-2s) = 1+3t-2s. $$ Substituting $$t = s$$ gives $$ -2(1-s) = 1+s $$ and thus $$s = 3$$. Therefore $$t = 3$$, yielding $$P = (7, 6, 8)$$ and $$Q = (5, 8, 12)$$.

The distance between these points is $$ PQ = \sqrt{(5-7)^2 + (8-6)^2 + (12-8)^2} = \sqrt{4+4+16} = \sqrt{24} = 2\sqrt{6}. $$ Hence, the correct answer is Option A: $$\boxed{2\sqrt{6}}$$.