The shortest distance between the lines $$\frac{x-5}{1} = \frac{y-2}{2} = \frac{z-4}{-3}$$ and $$\frac{x+3}{1} = \frac{y+5}{4} = \frac{z-1}{-5}$$ is

We need to find the shortest distance between the lines $$\frac{x-5}{1} = \frac{y-2}{2} = \frac{z-4}{-3}$$ and $$\frac{x+3}{1} = \frac{y+5}{4} = \frac{z-1}{-5}$$.

Identify points and direction vectors. Line 1: Point $$A(5, 2, 4)$$, direction $$\vec{b_1} = (1, 2, -3)$$.

Line 2: Point $$B(-3, -5, 1)$$, direction $$\vec{b_2} = (1, 4, -5)$$.

Find $$\vec{b_1} \times \vec{b_2}$$ by computing the determinant: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 1 & 4 & -5 \end{vmatrix}$$

$$= \hat{i}(2 \times(-5) - (-3) \times 4) - \hat{j}(1 \times(-5) - (-3) \times 1) + \hat{k}(1 \times 4 - 2 \times 1)$$

$$= \hat{i}(-10 + 12) - \hat{j}(-5 + 3) + \hat{k}(4 - 2) = 2\hat{i} + 2\hat{j} + 2\hat{k}$$

The magnitude of $$\vec{b_1} \times \vec{b_2}$$ is $$\sqrt{4 + 4 + 4} = 2\sqrt{3}$$.

The vector $$\vec{AB}$$ is given by $$\vec{AB} = B - A = (-3-5, -5-2, 1-4) = (-8, -7, -3)$$.

The shortest distance between the lines is given by $$d = \frac{|\vec{AB} \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$$.

$$\vec{AB} \cdot (\vec{b_1} \times \vec{b_2}) = (-8)(2) + (-7)(2) + (-3)(2) = -16 - 14 - 6 = -36$$

Substituting gives $$d = \frac{|-36|}{2\sqrt{3}} = \frac{36}{2\sqrt{3}} = \frac{18}{\sqrt{3}} = 6\sqrt{3}$$.

In conclusion, the shortest distance is $$6\sqrt{3}$$, which matches Option C.

Therefore, the answer is Option C.