Let the mirror image of the point $$(a, b, c)$$ with respect to the plane $$3x - 4y + 12z + 19 = 0$$ be $$(a - 6, \beta, \gamma)$$. If $$a + b + c = 5$$, then $$7\beta - 9\gamma$$ is equal to ______

The mirror image of $$(a, b, c)$$ in the plane $$3x - 4y + 12z + 19 = 0$$ is $$(a-6, \beta, \gamma)$$. Given $$a + b + c = 5$$, find $$7\beta - 9\gamma$$.

Using the mirror image formula for a point $$(x_0, y_0, z_0)$$ in the plane $$Ax + By + Cz + D = 0$$, we have

$$\frac{x - x_0}{A} = \frac{y - y_0}{B} = \frac{z - z_0}{C} = \frac{-2(Ax_0 + By_0 + Cz_0 + D)}{A^2 + B^2 + C^2}$$

Here $$A = 3, B = -4, C = 12, D = 19$$, and $$(x_0, y_0, z_0) = (a, b, c)$$. Since $$A^2 + B^2 + C^2 = 9 + 16 + 144 = 169$$, let

$$t = \frac{-2(3a - 4b + 12c + 19)}{169}$$.

Then the image is: $$(a + 3t, b - 4t, c + 12t) = (a - 6, \beta, \gamma)$$.

Comparing the $$x$$-coordinate gives $$a + 3t = a - 6 \Rightarrow 3t = -6 \Rightarrow t = -2$$.

Substituting this into the expression for $$t$$ yields $$\frac{-2(3a - 4b + 12c + 19)}{169} = -2$$, which implies $$3a - 4b + 12c + 19 = 169$$.

It follows that $$3a - 4b + 12c = 150$$ ... (i)

Also, $$a + b + c = 5$$ ... (ii)

Next, since $$\beta = b - 4t$$ we obtain

$$\beta = b - 4t = b - 4(-2) = b + 8$$

and since $$\gamma = c + 12t$$ we obtain

$$\gamma = c + 12t = c + 12(-2) = c - 24$$

Thus,

$$7\beta - 9\gamma = 7(b + 8) - 9(c - 24) = 7b + 56 - 9c + 216 = 7b - 9c + 272$$

To find $$7b - 9c$$, note that from (ii) we have $$a + b + c = 5$$, so $$a = 5 - b - c$$. Substituting this into (i) yields

$$3(5 - b - c) - 4b + 12c = 150$$

$$15 - 3b - 3c - 4b + 12c = 150$$

which gives $$-7b + 9c = 135$$ and hence $$7b - 9c = -135$$.

Therefore, $$7\beta - 9\gamma = -135 + 272 = 137$$.

The answer is $$\boxed{137}$$.