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?

$\text{Given: } \alpha,\beta,\gamma,\delta \in \mathbb{R},\ \alpha^2+\beta^2+\gamma^2 \neq 0,\ \alpha+\gamma = 1$ $\text{plane } \alpha x + \beta y + \gamma z = \delta$ $(3,2,-1) \text{ is the mirror image of } (1,0,-1)$ $\text{To find: which of (1)--(4) are TRUE}$ $1.\ \text{midpoint of } AB \text{ lies on the plane}$ $2.\ AB \parallel \text{ normal } (\alpha,\beta,\gamma)$ $3.\ \text{solve } \alpha,\beta,\gamma,\delta,\ \text{then check (1)--(4)}$ $M = \text{midpoint of } AB \text{ lies on the plane}$ $\vec{AB} \parallel (\alpha,\beta,\gamma)$ $M = \left(\dfrac{1+3}{2},\ \dfrac{0+2}{2},\ \dfrac{-1-1}{2}\right) = (2,\ 1,\ -1)$ $2\alpha + \beta - \gamma = \delta$ $\vec{AB} = (3-1,\ 2-0,\ -1-(-1)) = (2,\ 2,\ 0)$ $\dfrac{\alpha}{2} = \dfrac{\beta}{2} = \dfrac{\gamma}{0} \Rightarrow \gamma = 0,\ \alpha = \beta$ $\alpha + \gamma = 1,\ \gamma = 0 \Rightarrow \alpha = 1$ $\alpha = \beta = 1,\quad \gamma = 0$ $\text{plane}: x + y = \delta$ $2\alpha + \beta - \gamma = \delta \Rightarrow 2 + 1 - 0 = \delta$ $\delta = 3$ $\text{plane}: x + y = 3$ $\vec{n} = (1,1,0),\quad \text{plane } x + y = 3$ $\alpha + \beta = 1 + 1 = 2 \ \Rightarrow\ \text{(1) TRUE}$ $\delta - \gamma = 3 - 0 = 3 \ \Rightarrow\ \text{(2) TRUE}$ $\delta + \beta = 3 + 1 = 4 \ \Rightarrow\ \text{(3) TRUE}$ $\delta + \beta + \gamma = 3 + 1 + 0 = 4 \neq \delta = 3 \ \Rightarrow\ \text{(4) FALSE}$ $\boxed{\text{True: } (1),\ (2),\ (3)}$