Shortest distance between the lines $$\frac{x-1}{2} = \frac{y+8}{-7} = \frac{z-4}{5}$$ and $$\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3}$$ is:

Consider the two lines $$L_1: \frac{x-1}{2} = \frac{y+8}{-7} = \frac{z-4}{5}$$ passing through the point $$A(1, -8, 4)$$ with direction vector $$\vec{d_1} = (2, -7, 5)$$ and $$L_2: \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3}$$ passing through the point $$B(1, 2, 6)$$ with direction vector $$\vec{d_2} = (2, 1, -3)$$.

The cross product of the direction vectors is $$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = \hat{i}(21 - 5) - \hat{j}(-6 - 10) + \hat{k}(2 + 14) = (16, 16, 16),$$ and its magnitude is $$|\vec{d_1} \times \vec{d_2}| = 16\sqrt{3}.$$

The vector from A to B is $$\vec{AB} = B - A = (0, 10, 2).$$

The shortest distance between the lines is given by $$d = \frac{|\vec{AB} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} = \frac{|0\cdot16 + 10\cdot16 + 2\cdot16|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}.$$

Therefore, the correct answer is Option B: $$4\sqrt{3}$$.