Let $$(\alpha, \beta, \gamma)$$ be the image of the point $$(8, 5, 7)$$ in the line $$\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{5}$$. Then $$\alpha + \beta + \gamma$$ is equal to :

Find the image of point $$(8, 5, 7)$$ in the line $$\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{5}$$.

The line passes through $$A(1, -1, 2)$$ and has direction vector $$\vec{d} = (2, 3, 5)$$. Let the foot of the perpendicular from $$P(8, 5, 7)$$ to the line be $$F = (1+2t,\,-1+3t,\;2+5t)$$, so that $$\vec{PF} = (2t-7,\;3t-6,\;5t-5)$$.

Enforcing $$\vec{PF}\cdot\vec{d}=0$$ yields $$ 2(2t-7) + 3(3t-6) + 5(5t-5) = 0 $$ $$ 4t - 14 + 9t - 18 + 25t - 25 = 0 $$ $$ 38t - 57 = 0 $$ and hence $$ t = \frac{3}{2} $$.

Substituting back gives $$F = (1+3,\,-1+4.5,\;2+7.5) = (4,\;3.5,\;9.5)$$. Since $$F$$ is the midpoint of $$P$$ and its image $$(\alpha,\beta,\gamma)$$, we get $$ \alpha = 2(4) - 8 = 0, \quad \beta = 2(3.5) - 5 = 2, \quad \gamma = 2(9.5) - 7 = 12. $$

Therefore, $$ \alpha + \beta + \gamma = 0 + 2 + 12 = 14 $$. The correct answer is Option (3): 14.