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?

$\text{Given: } \alpha, \beta, \gamma, \delta \in \mathbb{R},\quad \alpha^2 + \beta^2 + \gamma^2 \ne 0,\quad \alpha + \gamma = 1$ $\text{Mirror image of } P(1, 0, -1) \text{ is } Q(3, 2, -1) \text{ in plane } \alpha x + \beta y + \gamma z = \delta$ $\text{To find: which of the following are TRUE?}$ $(1)\ \alpha + \beta = 2 \qquad (2)\ \delta - \gamma = 3$ $(3)\ \delta + \beta = 4 \qquad (4)\ \delta + \beta + \gamma = \delta$ $P(1,0,-1) \quad \text{and} \quad Q(3,2,-1)$ $\vec{PQ} = Q - P = (2,\ 2,\ 0)$ $M = \dfrac{P+Q}{2} = (2,\ 1,\ -1)$ $\text{Plan: } 1.\ (\alpha, \beta, \gamma) \parallel \vec{PQ}$ $2.\ \alpha + \gamma = 1 \Rightarrow \alpha, \beta, \gamma$ $3.\ M \text{ on plane} \Rightarrow \delta,\ \text{then check options}$ $(\alpha, \beta, \gamma) \parallel \vec{PQ} = (2, 2, 0)$ $\alpha : \beta : \gamma = 2 : 2 : 0 \Rightarrow \gamma = 0,\ \alpha = \beta$ $\alpha + \gamma = 1 \Rightarrow \alpha = 1,\ \beta = 1,\ \gamma = 0$ $\alpha x + \beta y + \gamma z = \delta \Rightarrow x + y = \delta$ $M(2,1,-1) \in x + y = \delta \Rightarrow 2 + 1 = \delta$ $\delta = 3$ $(1)\ \alpha + \beta = 1 + 1 = 2 \quad ✓$ $(2)\ \delta - \gamma = 3 - 0 = 3 \quad ✓$ $(3)\ \delta + \beta = 3 + 1 = 4 \quad ✓$ $(4)\ \delta + \beta + \gamma = 3 + 1 + 0 = 4 \ne \delta = 3 \quad ✗$ $\therefore \alpha = \beta = 1,\quad \gamma = 0,\quad \delta = 3$ $\boxed{(1),\ (2),\ (3) \text{ are TRUE}}$