Let $$(\alpha, \beta, \gamma)$$ be the image of point $$P(2, 3, 5)$$ in the plane $$2x + y - 3z = 6$$. Then $$\alpha + \beta + \gamma$$ is equal to

To find the image of $$P(2, 3, 5)$$ in the plane $$2x + y - 3z = 6$$.

The formula for the image $$(\alpha, \beta, \gamma)$$ of point $$(x_1, y_1, z_1)$$ in plane $$ax + by + cz = d$$ is:

$$\frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = \frac{\gamma - z_1}{c} = \frac{-2(ax_1 + by_1 + cz_1 - d)}{a^2 + b^2 + c^2}$$

$$\frac{-2(2(2) + 1(3) - 3(5) - 6)}{4 + 1 + 9} = \frac{-2(4 + 3 - 15 - 6)}{14} = \frac{-2(-14)}{14} = 2$$

Therefore:

$$\alpha = 2 + 2(2) = 6$$

$$\beta = 3 + 2(1) = 5$$

$$\gamma = 5 + 2(-3) = -1$$

$$\alpha + \beta + \gamma = 6 + 5 + (-1) = 10$$