The shortest distance between the lines $$\frac{x-3}{2} = \frac{y-2}{3} = \frac{z-1}{-1}$$ and $$\frac{x+3}{2} = \frac{y-6}{1} = \frac{z-5}{3}$$ is

We need to find the shortest distance between the lines:

$$L_1: \frac{x-3}{2} = \frac{y-2}{3} = \frac{z-1}{-1}$$

$$L_2: \frac{x+3}{2} = \frac{y-6}{1} = \frac{z-5}{3}$$

Point on $$L_1$$: $$\vec{a_1} = (3, 2, 1)$$, direction: $$\vec{b_1} = (2, 3, -1)$$

Point on $$L_2$$: $$\vec{a_2} = (-3, 6, 5)$$, direction: $$\vec{b_2} = (2, 1, 3)$$

$$\vec{a_2} - \vec{a_1} = (-6, 4, 4)$$

$$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 3 & -1 \\ 2 & 1 & 3 \end{vmatrix}$$

$$= \vec{i}(9 + 1) - \vec{j}(6 + 2) + \vec{k}(2 - 6)$$

$$= (10, -8, -4)$$

$$|\vec{b_1} \times \vec{b_2}| = \sqrt{100 + 64 + 16} = \sqrt{180} = 6\sqrt{5}$$

Shortest distance:

$$d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$$

$$= \frac{|(-6)(10) + (4)(-8) + (4)(-4)|}{6\sqrt{5}}$$

$$= \frac{|-60 - 32 - 16|}{6\sqrt{5}}$$

$$= \frac{108}{6\sqrt{5}} = \frac{18}{\sqrt{5}}$$

The correct answer is Option A: $$\frac{18}{\sqrt{5}}$$.