Let the plane $$2x + 3y + z + 20 = 0$$ be rotated through a right angle about its line of intersection with the plane $$x - 3y + 5z = 8$$. If the mirror image of the point $$(2, -\frac{1}{2}, 2)$$ in the rotated plane is $$B(a, b, c)$$, then

The plane $$P_1: 2x + 3y + z + 20 = 0$$ is rotated through a right angle about its line of intersection with $$P_2: x - 3y + 5z - 8 = 0$$.

Family of planes through the line of intersection.

The rotated plane belongs to the family: $$P_1 + \lambda P_2 = 0$$

$$(2 + \lambda)x + (3 - 3\lambda)y + (1 + 5\lambda)z + (20 - 8\lambda) = 0$$

The rotated plane is perpendicular to $$P_1$$.

Normal of $$P_1$$: $$\vec{n_1} = (2, 3, 1)$$

Normal of rotated plane: $$\vec{n} = (2+\lambda, 3-3\lambda, 1+5\lambda)$$

$$\vec{n} \cdot \vec{n_1} = 0$$

$$2(2+\lambda) + 3(3-3\lambda) + 1(1+5\lambda) = 0$$

$$4 + 2\lambda + 9 - 9\lambda + 1 + 5\lambda = 0$$

$$14 - 2\lambda = 0 \Rightarrow \lambda = 7$$

The rotated plane is:

$$9x - 18y + 36z - 36 = 0 \Rightarrow x - 2y + 4z - 4 = 0$$

Find the mirror image of $$(2, -1/2, 2)$$ in this plane.

The formula for the mirror image of point $$(x_0, y_0, z_0)$$ in plane $$ax + by + cz + d = 0$$:

$$\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} = \frac{-2(ax_0 + by_0 + cz_0 + d)}{a^2 + b^2 + c^2}$$

Here $$a = 1, b = -2, c = 4, d = -4$$ and $$(x_0, y_0, z_0) = (2, -1/2, 2)$$.

$$ax_0 + by_0 + cz_0 + d = 2 + 1 + 8 - 4 = 7$$

$$a^2 + b^2 + c^2 = 1 + 4 + 16 = 21$$

$$t = \frac{-2(7)}{21} = \frac{-2}{3}$$

$$a = x = 2 + 1 \cdot (-2/3) = 4/3$$

$$b = y = -1/2 + (-2)(-2/3) = -1/2 + 4/3 = 5/6$$

$$c = z = 2 + 4(-2/3) = 2 - 8/3 = -2/3$$

Check which option matches.

$$B = (4/3, 5/6, -2/3)$$

$$\frac{a}{8} = \frac{4/3}{8} = \frac{1}{6}$$

$$\frac{b}{5} = \frac{5/6}{5} = \frac{1}{6}$$

$$\frac{c}{-4} = \frac{-2/3}{-4} = \frac{1}{6}$$

All ratios equal $$\frac{1}{6}$$ $$\checkmark$$

The correct answer is Option A: $$\frac{a}{8} = \frac{b}{5} = \frac{c}{-4}$$.