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

 First, we observe that the line passes through $$A(1, 2, 1)$$ with direction vector $$\vec{d} = (2, 1, 3)$$, so a general point on the line can be written as $$Q = (1 + 2t,\; 2 + t,\; 1 + 3t)$$.

Next, we form the vector from this point to $$P$$, namely $$\overrightarrow{QP} = P - Q = (4 - 1 - 2t,\; 4 - 2 - t,\; 3 - 1 - 3t) = (3 - 2t,\; 2 - t,\; 2 - 3t)$$. 

Since the foot of the perpendicular from $$P$$ to the line occurs when this vector is orthogonal to the direction vector $$\vec{d}$$, we impose $$\overrightarrow{QP}\cdot \vec{d} = 0$$, which gives $$2(3 - 2t) + 1(2 - t) + 3(2 - 3t) = 0,$$ leading to $$6 - 4t + 2 - t + 6 - 9t = 14 - 14t = 0$$ and hence $$t = 1$$.

Substituting $$t = 1$$ back into the parametric form yields the foot of the perpendicular $$F = (1 + 2(1),\; 2 + 1,\; 1 + 3(1)) = (3, 3, 4).$$

Finally, the reflection of $$P$$ across this foot is calculated by $$(\alpha, \beta, \gamma) = 2F - P = 2(3, 3, 4) - (4, 4, 3) = (6 - 4,\; 6 - 4,\; 8 - 3) = (2, 2, 5),$$ and thus $$\alpha + \beta + \gamma = 2 + 2 + 5 = 9.$$