Let $$P(\alpha, \beta, \gamma)$$ be the image of the point $$Q(1, 6, 4)$$ in the line $$\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$$. Then $$2\alpha + \beta + \gamma$$ is equal to _____

We need to find the image $$P(\alpha,\beta,\gamma)$$ of the point $$Q(1,6,4)$$ in the line $$\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$$. The line has direction ratios $$(1,2,3)$$ and passes through $$(0,1,2)$$.

First we find the foot $$F$$ of the perpendicular from $$Q$$ to the line. A general point on the line is $$(t,1+2t,2+3t)$$. The vector from this point to $$Q$$ is $$(1-t,6-1-2t,4-2-3t) = (1-t,5-2t,2-3t)$$. For perpendicularity to $$(1,2,3)$$ we set

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

$$ 1-t + 10-4t + 6-9t = 0 $$

$$ 17 - 14t = 0 \implies t = \frac{17}{14}. $$

Substituting back gives the foot

$$ F = \Bigl(\frac{17}{14},\,1+\frac{34}{14},\,2+\frac{51}{14}\Bigr) = \Bigl(\frac{17}{14},\,\frac{48}{14},\,\frac{79}{14}\Bigr). $$

Next the image $$P$$ is given by $$P = 2F - Q$$. Hence,

$$ \alpha = 2 \cdot \frac{17}{14} - 1 = \frac{34}{14} - 1 = \frac{20}{14} = \frac{10}{7}, $$

$$ \beta = 2 \cdot \frac{48}{14} - 6 = \frac{96}{14} - 6 = \frac{96-84}{14} = \frac{12}{14} = \frac{6}{7}, $$

$$ \gamma = 2 \cdot \frac{79}{14} - 4 = \frac{158}{14} - 4 = \frac{158-56}{14} = \frac{102}{14} = \frac{51}{7}. $$

Therefore,

$$ 2\alpha + \beta + \gamma = \frac{20}{7} + \frac{6}{7} + \frac{51}{7} = \frac{77}{7} = 11. $$

The answer is $$\boxed{11}$$.