The distance of the point $$(7, 10, 11)$$ from the line $$\dfrac{x - 4}{1} = \dfrac{y - 4}{0} = \dfrac{z - 2}{3}$$ along the line $$\dfrac{x - 9}{2} = \dfrac{y - 13}{3} = \dfrac{z - 17}{6}$$ is

We need to find the distance of the point $$P(7, 10, 11)$$ from line $$L_1: \dfrac{x-4}{1} = \dfrac{y-4}{0} = \dfrac{z-2}{3}$$ along line $$L_2: \dfrac{x-9}{2} = \dfrac{y-13}{3} = \dfrac{z-17}{6}$$.

The line through $$P(7, 10, 11)$$ parallel to $$L_2$$ (direction $$(2, 3, 6)$$) is: $$\dfrac{x-7}{2} = \dfrac{y-10}{3} = \dfrac{z-11}{6} = t$$

A general point on this line is $$(7 + 2t, 10 + 3t, 11 + 6t)$$.

This point must lie on $$L_1$$. From $$L_1$$: $$y = 4$$ (since the direction ratio for $$y$$ is 0).

So $$10 + 3t = 4$$, giving $$t = -2$$.

The point of intersection is $$(7 - 4, 4, 11 - 12) = (3, 4, -1)$$.

Let us verify this lies on $$L_1$$: $$\dfrac{3-4}{1} = -1$$ and $$\dfrac{-1-2}{3} = -1$$. Yes, it does.

The distance from $$P(7, 10, 11)$$ to $$(3, 4, -1)$$ is:

$$d = \sqrt{(7-3)^2 + (10-4)^2 + (11-(-1))^2} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$$

Hence, the correct answer is Option B.