The distance, of the point $$(7, -2, 11)$$ from the line $$\frac{x-6}{1} = \frac{y-4}{0} = \frac{z-8}{3}$$ along the line $$\frac{x-5}{2} = \frac{y-1}{-3} = \frac{z-5}{6}$$, is :

We need the distance from $$(7, -2, 11)$$ to the line $$L_1: \frac{x-6}{1} = \frac{y-4}{0} = \frac{z-8}{3}$$ along the line $$L_2: \frac{x-5}{2} = \frac{y-1}{-3} = \frac{z-5}{6}$$.

A general point on $$L_2$$ through $$(7, -2, 11)$$ (with parameter $$\mu$$):

The direction of $$L_2$$ is $$(2, -3, 6)$$. Starting from $$(7, -2, 11)$$:

Point: $$(7 + 2\mu, -2 - 3\mu, 11 + 6\mu)$$

This point should lie on $$L_1$$: $$\frac{x-6}{1} = \frac{y-4}{0} = \frac{z-8}{3}$$.

From $$\frac{y-4}{0}$$: we need $$y = 4$$, so $$-2 - 3\mu = 4$$, giving $$\mu = -2$$.

Check: $$x = 7 + 2(-2) = 3$$, $$y = -2 - 3(-2) = 4$$, $$z = 11 + 6(-2) = -1$$.

Verify on $$L_1$$: $$\frac{3-6}{1} = -3$$ and $$\frac{-1-8}{3} = -3$$. ✓

Distance = $$|\mu| \times |L_2 \text{ direction}| = 2 \times \sqrt{4 + 9 + 36} = 2 \times 7 = 14$$

The answer is $$\boxed{14}$$, which corresponds to Option (2).