If the equation of plane passing through the mirror image of a point $$(2, 3, 1)$$ with respect to line $$\frac{x+1}{2} = \frac{y-3}{1} = \frac{z+2}{-1}$$ and containing the line $$\frac{x-2}{3} = \frac{1-y}{2} = \frac{z+1}{1}$$ is $$\alpha x + \beta y + \gamma z = 24$$ then $$\alpha + \beta + \gamma$$ is equal to:

We need the mirror image of $$(2, 3, 1)$$ in the line $$\frac{x+1}{2} = \frac{y-3}{1} = \frac{z+2}{-1}$$. The line passes through $$A = (-1, 3, -2)$$ with direction $$\vec{d} = (2, 1, -1)$$.

Let $$P = (2, 3, 1)$$. The vector $$\vec{AP} = (2-(-1),\; 3-3,\; 1-(-2)) = (3, 0, 3)$$.

The foot of the perpendicular $$F$$ from $$P$$ onto the line is found by projecting $$\vec{AP}$$ onto $$\vec{d}$$: $$t = \frac{\vec{AP} \cdot \vec{d}}{|\vec{d}|^2} = \frac{3(2) + 0(1) + 3(-1)}{4+1+1} = \frac{3}{6} = \frac{1}{2}$$.

So $$F = A + t\vec{d} = (-1+1,\; 3+\tfrac{1}{2},\; -2-\tfrac{1}{2}) = (0, \tfrac{7}{2}, -\tfrac{5}{2})$$.

The mirror image $$P'$$ satisfies $$F = \frac{P + P'}{2}$$, giving $$P' = 2F - P = (0-2,\; 7-3,\; -5-1) = (-2, 4, -6)$$.

The plane contains the line $$\frac{x-2}{3} = \frac{1-y}{2} = \frac{z+1}{1}$$, which passes through $$(2, 1, -1)$$ with direction $$(3, -2, 1)$$, and also passes through $$P' = (-2, 4, -6)$$.

A second direction vector in the plane is $$\vec{v} = P' - (2,1,-1) = (-4, 3, -5)$$.

The normal to the plane is $$\vec{n} = (3,-2,1) \times (-4,3,-5)$$. Computing: $$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 1 \\ -4 & 3 & -5 \end{vmatrix} = \hat{i}(10-3) - \hat{j}(-15+4) + \hat{k}(9-8) = (7, 11, 1)$$.

The plane equation through $$(2,1,-1)$$ is $$7(x-2) + 11(y-1) + 1(z+1) = 0$$, which simplifies to $$7x + 11y + z = 24$$.

Comparing with $$\alpha x + \beta y + \gamma z = 24$$, we get $$\alpha = 7$$, $$\beta = 11$$, $$\gamma = 1$$, so $$\alpha + \beta + \gamma = 19$$.