Let $$(\alpha, \beta, \gamma)$$ be mirror image of the point $$(2, 3, 5)$$ in the line $$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$$. Then $$2\alpha + 3\beta + 4\gamma$$ is equal to

Given: Point $$P(2, 3, 5)$$, line equation: $$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} = \lambda$$

Let $$F$$ be the foot of the perpendicular from $$P$$ onto the line: $$F = (2\lambda + 1, \ 3\lambda + 2, \ 4\lambda + 3)$$

Calculating the direction ratios of $$PF$$:

$$\vec{PF} = (2\lambda + 1 - 2)\hat{i} + (3\lambda + 2 - 3)\hat{j} + (4\lambda + 3 - 5)\hat{k} = (2\lambda - 1)\hat{i} + (3\lambda - 1)\hat{j} + (4\lambda - 2)\hat{k}$$

Since $$PF$$ is perpendicular to the line (direction ratios $$2, 3, 4$$):

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

$$4\lambda - 2 + 9\lambda - 3 + 16\lambda - 8 = 0 \implies 29\lambda = 13 \implies \lambda = \frac{13}{29}$$

$$F = \left(\frac{55}{29}, \ \frac{97}{29}, \ \frac{139}{29}\right)$$

Using the midpoint formula where $$F$$ is the midpoint of $$P(2,3,5)$$ and its image $$P'(\alpha, \beta, \gamma)$$:

$$\frac{2 + \alpha}{2} = \frac{55}{29} \implies \alpha = \frac{110}{29} - 2 = \frac{52}{29}$$

$$\frac{3 + \beta}{2} = \frac{97}{29} \implies \beta = \frac{194}{29} - 3 = \frac{107}{29}$$

$$\frac{5 + \gamma}{2} = \frac{139}{29} \implies \gamma = \frac{278}{29} - 5 = \frac{133}{29}$$

$$2\alpha + 3\beta + 4\gamma = \frac{2(52) + 3(107) + 4(133)}{29}$$

$$2\alpha + 3\beta + 4\gamma = \frac{104 + 321 + 532}{29} = \frac{957}{29} = 33$$