Let the foot of the perpendicular from the point $$(1, 2, 4)$$ on the line $$\frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3}$$ be $$P$$. Then the distance of $$P$$ from the plane $$3x + 4y + 12z + 23 = 0$$ is

We need to find the foot of perpendicular $$P$$ from $$(1, 2, 4)$$ to the line $$\frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3} = \lambda$$.

A general point on the line is $$(4\lambda - 2, 2\lambda + 1, 3\lambda - 1)$$.

The direction ratios of the line are $$(4, 2, 3)$$.

The vector from $$(1, 2, 4)$$ to the general point is:

$$(4\lambda - 3, 2\lambda - 1, 3\lambda - 5)$$

For the foot of perpendicular, this vector must be perpendicular to the direction of the line:

$$4(4\lambda - 3) + 2(2\lambda - 1) + 3(3\lambda - 5) = 0$$

$$16\lambda - 12 + 4\lambda - 2 + 9\lambda - 15 = 0$$

$$29\lambda - 29 = 0 \implies \lambda = 1$$

So $$P = (4(1) - 2, 2(1) + 1, 3(1) - 1) = (2, 3, 2)$$.

Now we find the distance of $$P(2, 3, 2)$$ from the plane $$3x + 4y + 12z + 23 = 0$$:

$$d = \frac{|3(2) + 4(3) + 12(2) + 23|}{\sqrt{9 + 16 + 144}}$$

$$= \frac{|6 + 12 + 24 + 23|}{\sqrt{169}}$$

$$= \frac{65}{13} = 5$$

The correct answer is Option C: $$\frac{65}{13}$$.