If the shortest distance between the lines $$\frac{x-4}{1} = \frac{y+1}{2} = \frac{z}{-3}$$ and $$\frac{x-\lambda}{2} = \frac{y+1}{4} = \frac{z-2}{-5}$$ is $$\frac{6}{\sqrt{5}}$$, then the sum of all possible values of $$\lambda$$ is :

Line 1: $$\frac{x-4}{1} = \frac{y+1}{2} = \frac{z}{-3}$$, point $$A(4, -1, 0)$$, direction $$\vec{d_1} = (1, 2, -3)$$.

Line 2: $$\frac{x-\lambda}{2} = \frac{y+1}{4} = \frac{z-2}{-5}$$, point $$B(\lambda, -1, 2)$$, direction $$\vec{d_2} = (2, 4, -5)$$.

$$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} = \hat{i}(-10+12) - \hat{j}(-5+6) + \hat{k}(4-4) = (2, -1, 0)$$

$$|\vec{d_1} \times \vec{d_2}| = \sqrt{4 + 1 + 0} = \sqrt{5}$$

$$\vec{AB} = (\lambda - 4, 0, 2)$$

Shortest distance = $$\frac{|\vec{AB} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} = \frac{|2(\lambda - 4) + 0 + 0|}{\sqrt{5}} = \frac{2|\lambda - 4|}{\sqrt{5}}$$

Setting this equal to $$\frac{6}{\sqrt{5}}$$:

$$2|\lambda - 4| = 6$$

$$|\lambda - 4| = 3$$

$$\lambda = 7$$ or $$\lambda = 1$$

Sum of all possible values: $$7 + 1 = 8$$.

The answer is $$\boxed{8}$$, which corresponds to Option (2).