The square of the distance of the image of the point $$(6, 1, 5)$$ in the line $$\frac{x-1}{3} = \frac{y}{2} = \frac{z-2}{4}$$, from the origin is ________

We need to find the square of the distance from the origin to the image of the point $$(6, 1, 5)$$ in the line $$\frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4}$$.

A general point on the line is:

$$P = (1 + 3t, \; 2t, \; 2 + 4t)$$

The direction from the given point $$(6, 1, 5)$$ to $$P$$ is:

$$(1 + 3t - 6, \; 2t - 1, \; 2 + 4t - 5) = (3t - 5, \; 2t - 1, \; 4t - 3)$$

For the perpendicular from the point to the line, this direction must be perpendicular to the line's direction vector $$(3, 2, 4)$$:

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

$$9t - 15 + 4t - 2 + 16t - 12 = 0$$

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

Substituting $$t = 1$$:

$$\text{Foot} = (1 + 3, \; 2, \; 2 + 4) = (4, 2, 6)$$

The image is the reflection of $$(6, 1, 5)$$ through the foot $$(4, 2, 6)$$. Since the foot is the midpoint of the original point and its image:

$$\text{Image} = (2 \times 4 - 6, \; 2 \times 2 - 1, \; 2 \times 6 - 5) = (2, 3, 7)$$

$$d^2 = 2^2 + 3^2 + 7^2 = 4 + 9 + 49 = 62$$

The answer is $$\textbf{62}$$.