If the foot of the perpendicular from point $$(4, 3, 8)$$ on the line $$L_1: \frac{x-a}{l} = \frac{y-3}{3} = \frac{z-b}{4}$$, $$l \neq 0$$ is $$(3, 5, 7)$$, then the shortest distance between the line $$L_1$$ and line $$L_2: \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}$$ is equal to:

We are given the point $$(4, 3, 8)$$ and its foot of perpendicular $$(3, 5, 7)$$ on line $$L_1$$. The direction vector of the perpendicular from $$(4,3,8)$$ to $$(3,5,7)$$ is $$(3-4, 5-3, 7-8) = (-1, 2, -1)$$.

Since $$(3, 5, 7)$$ lies on $$L_1$$, we get $$\frac{3-a}{l} = \frac{5-3}{3} = \frac{7-b}{4}$$. From the middle ratio, $$\frac{2}{3}$$. So $$\frac{7-b}{4} = \frac{2}{3}$$, giving $$7-b = \frac{8}{3}$$, hence $$b = \frac{13}{3}$$. Also $$\frac{3-a}{l} = \frac{2}{3}$$, so $$3-a = \frac{2l}{3}$$.

The direction vector of $$L_1$$ is $$(l, 3, 4)$$. Since the perpendicular direction $$(-1, 2, -1)$$ is orthogonal to $$(l, 3, 4)$$, we have $$-l + 6 - 4 = 0$$, giving $$l = 2$$.

So the direction vector of $$L_1$$ is $$(2, 3, 4)$$ and from $$3-a = \frac{2(2)}{3} = \frac{4}{3}$$, we get $$a = \frac{5}{3}$$. Thus $$L_1$$ passes through $$(3, 5, 7)$$ with direction $$(2, 3, 4)$$.

Line $$L_2$$ passes through $$(2, 4, 5)$$ with direction $$(3, 4, 5)$$. The vector joining points on $$L_1$$ and $$L_2$$ is $$(3-2, 5-4, 7-5) = (1, 1, 2)$$.

The cross product of the direction vectors $$\vec{d_1} = (2, 3, 4)$$ and $$\vec{d_2} = (3, 4, 5)$$ is: $$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9) = (-1, 2, -1)$$

The magnitude is $$|\vec{d_1} \times \vec{d_2}| = \sqrt{1+4+1} = \sqrt{6}$$.

The shortest distance is $$\frac{|(1, 1, 2) \cdot (-1, 2, -1)|}{\sqrt{6}} = \frac{|-1+2-2|}{\sqrt{6}} = \frac{1}{\sqrt{6}}$$.

The answer is $$\frac{1}{\sqrt{6}}$$, which corresponds to Option (2).