Let $\alpha, \beta, \gamma, \delta$ be real numbers such that $\alpha^2 + \beta^2 + \gamma^2 \neq 0$ and $\alpha + \gamma = 1$. Suppose the point $(3, 2, -1)$ is the mirror image of the point $(1, 0, -1)$ with respect to the plane $\alpha x + \beta y + \gamma z = \delta$. Then which of the following statements is/are TRUE?
Step-by-Step Solution
Key Concept: The mirror image condition implies the midpoint lies on the plane, and the line connecting the points is perpendicular to the plane (i.e., parallel to the normal vector).
1. **Midpoint on Plane**: The midpoint $ M $ of $ (1, 0, -1) $ and $ (3, 2, -1) $ is $ (2, 1, -1) $. Substituting into the plane equation:
$$
2\alpha + \beta - \gamma = \delta \quad \text{(Equation 1)}.
$$
2. **Normal Vector Alignment**: The line joining the points has direction vector $ (2, 2, 0) $, which must align with the plane’s normal vector $ (\alpha, \beta, \gamma) $. Thus:
$$
\alpha = 2k, \quad \beta = 2k, \quad \gamma = 0 \quad \text{for some } k \in \mathbb{R}.
$$
3. **Solve for Constants**: Using $ \alpha + \gamma = 1 $:
$$
2k + 0 = 1 \implies k = \frac{1}{2} \implies \alpha = 1, \beta = 1, \gamma = 0.
$$
Substitute into Equation 1:
$$
2(1) + 1 - 0 = \delta \implies \delta = 3.
$$
4. **Final Plane Equation**: The plane is $ x + y = 3 $, with normal vector $ (1, 1, 0) $.
Correct Answer: A, B, C