If the mirror image of the point $$P(3, 4, 9)$$ in the line $$\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1}$$ is $$(\alpha, \beta, \gamma)$$, then $$14(\alpha + \beta + \gamma)$$ is:

We need to find the mirror image of $$P(3, 4, 9)$$ in the line $$\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1}$$.

Since we need the foot of the perpendicular from P to the line, we take a point on the line, $$A(1, -1, 2)$$, with direction vector $$\vec{d} = (3, 2, 1)$$.

A general point on the line is $$(1 + 3t,\,-1 + 2t,\,2 + t)$$, so the vector from this point to P is $$(3 - (1 + 3t),\,4 - (-1 + 2t),\,9 - (2 + t)) = (2 - 3t,\,5 - 2t,\,7 - t)\,.$$

For the foot of perpendicular, this vector must be perpendicular to $$\vec{d}$$, so

$$ (2 - 3t)(3) + (5 - 2t)(2) + (7 - t)(1) = 0 $$

$$ 6 - 9t + 10 - 4t + 7 - t = 0 $$

$$ 23 - 14t = 0 \implies t = \frac{23}{14}\,.$$

Substituting back gives

$$ N = \Bigl(1 + 3\cdot\frac{23}{14},\,-1 + 2\cdot\frac{23}{14},\,2 + \frac{23}{14}\Bigr) = \Bigl(\frac{83}{14},\,\frac{32}{14},\,\frac{51}{14}\Bigr) = \Bigl(\frac{83}{14},\,\frac{16}{7},\,\frac{51}{14}\Bigr)\,.$$

Now, since N is the midpoint of P and its mirror image Q, we set

$$ \alpha = 2\cdot\frac{83}{14} - 3 = \frac{166}{14} - \frac{42}{14} = \frac{124}{14} = \frac{62}{7}, $$

$$ \beta = 2\cdot\frac{16}{7} - 4 = \frac{32}{7} - \frac{28}{7} = \frac{4}{7}, $$

$$ \gamma = 2\cdot\frac{51}{14} - 9 = \frac{102}{14} - \frac{126}{14} = -\frac{24}{14} = -\frac{12}{7}. $$

Therefore,

$$ \alpha + \beta + \gamma = \frac{62}{7} + \frac{4}{7} - \frac{12}{7} = \frac{54}{7}, $$

and

$$ 14(\alpha + \beta + \gamma) = 14 \times \frac{54}{7} = 2 \times 54 = 108. $$

The correct answer is Option 3: 108.