Let the image of the point $$P(1, 2, 3)$$ in the line $$L : \frac{x-6}{3} = \frac{y-1}{2} = \frac{z-2}{3}$$ be $$Q$$. Let $$R(\alpha, \beta, \gamma)$$ be a point that divides internally the line segment $$PQ$$ in the ratio 1 : 3. Then the value of $$22(\alpha + \beta + \gamma)$$ is equal to

We need to find the image $$Q$$ of $$P(1,2,3)$$ in the line $$L: \frac{x-6}{3} = \frac{y-1}{2} = \frac{z-2}{3}$$, then find point $$R$$ dividing $$PQ$$ in ratio 1:3 internally. A general point on $$L$$ is $$(6+3t, 1+2t, 2+3t)$$, and the direction of $$L$$ is $$(3,2,3)$$. The vector from $$P(1,2,3)$$ to the point on $$L$$ is $$(5+3t, -1+2t, -1+3t)$$, and for perpendicularity with the line direction we require $$3(5+3t) + 2(-1+2t) + 3(-1+3t) = 0$$, which simplifies to $$15 + 9t - 2 + 4t - 3 + 9t = 0$$, yielding $$22t + 10 = 0$$ and hence $$t = -\frac{10}{22} = -\frac{5}{11}$$.

Substituting this value of $$t$$ into the coordinates of the line gives the foot of the perpendicular $$M = \left(6 - \frac{15}{11},\ 1 - \frac{10}{11},\ 2 - \frac{15}{11}\right) = \left(\frac{51}{11},\ \frac{1}{11},\ \frac{7}{11}\right)$$.

Since $$M$$ is the midpoint of $$PQ$$, the image $$Q$$ of $$P$$ in the line is $$Q = 2M - P = \left(\frac{102}{11} - 1,\ \frac{2}{11} - 2,\ \frac{14}{11} - 3\right) = \left(\frac{91}{11},\ \frac{-20}{11},\ \frac{-19}{11}\right)$$.

Point $$R$$ dividing $$PQ$$ in the ratio 1:3 internally is given by $$R = \frac{3P + 1Q}{4} = \frac{1}{4}\left(3P + Q\right)$$, with $$\alpha = \frac{3(1) + \frac{91}{11}}{4} = \frac{\frac{33 + 91}{11}}{4} = \frac{124}{44} = \frac{31}{11},$$ $$\beta = \frac{3(2) + \frac{-20}{11}}{4} = \frac{\frac{66 - 20}{11}}{4} = \frac{46}{44} = \frac{23}{22},$$ $$\gamma = \frac{3(3) + \frac{-19}{11}}{4} = \frac{\frac{99 - 19}{11}}{4} = \frac{80}{44} = \frac{20}{11}.$$

Hence $$\alpha + \beta + \gamma = \frac{31}{11} + \frac{23}{22} + \frac{20}{11} = \frac{62}{22} + \frac{23}{22} + \frac{40}{22} = \frac{125}{22}$$, and so $$22(\alpha + \beta + \gamma) = 22 \cdot \frac{125}{22} = 125$$.

The answer is $$\boxed{125}$$.