Let $$P$$ be the plane passing through the line $$\frac{x-1}{1} = \frac{y-2}{-3} = \frac{z+5}{7}$$ and the point $$(2, 4, -3)$$. If the image of the point $$(-1, 3, 4)$$ in the plane $$P$$ is $$(\alpha, \beta, \gamma)$$, then $$\alpha + \beta + \gamma$$ is equal to

The line is $$\frac{x-1}{1} = \frac{y-2}{-3} = \frac{z+5}{7}$$ with direction ratios $$(1, -3, 7)$$ and point $$(1, 2, -5)$$.

The plane passes through this line and the point $$(2, 4, -3)$$.

First, we find the equation of the plane.

Vector from $$(1,2,-5)$$ to $$(2,4,-3)$$: $$(1, 2, 2)$$

Direction of line: $$(1, -3, 7)$$

Normal to plane = $$(1, 2, 2) \times (1, -3, 7)$$:

$$= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 1 & -3 & 7 \end{vmatrix} = \hat{i}(14+6) - \hat{j}(7-2) + \hat{k}(-3-2)$$

$$= (20, -5, -5) = 5(4, -1, -1)$$

Plane equation: $$4(x-1) - 1(y-2) - 1(z+5) = 0$$

$$4x - 4 - y + 2 - z - 5 = 0$$

$$4x - y - z = 7$$

Next, we find image of $$(-1, 3, 4)$$ in plane $$4x - y - z = 7$$.

Using the image formula: $$\frac{\alpha+1}{4} = \frac{\beta-3}{-1} = \frac{\gamma-4}{-1} = \frac{-2(4(-1)-3-4-7)}{16+1+1}$$

$$= \frac{-2(-4-3-4-7)}{18} = \frac{-2(-18)}{18} = \frac{36}{18} = 2$$

$$\alpha = -1 + 8 = 7$$

$$\beta = 3 - 2 = 1$$

$$\gamma = 4 - 2 = 2$$

$$\alpha + \beta + \gamma = 7 + 1 + 2 = 10$$

The correct answer is Option 1: 10.