Let α, β, γ, be real numbers such that α ^ 2 + β ^ 2 + γ ^ 2 ≠ 0...

From the diagram,

Mid point of A & A ' lies on the plane, i.e., $B\left(2, 1, - 1\right)$

$2\alpha + \beta - \gamma = \delta \ldots \left(1\right)$

$\alpha + \gamma = 1 \ldots . \left(2\right)$

Direction Ratio of ${AA}^{'}\left(2, 2, 0\right)$ will be proportional to normal to the plane,

$\dfrac{\alpha}{2} = \dfrac{\beta}{2} = \dfrac{\gamma}{0} = \lambda$

$\alpha = 2\lambda, \beta = 2\lambda, \gamma = 0$

$∴ 2\lambda + 0 = 1 ⇒ \lambda = \dfrac{1}{2}$

$∴ \alpha = 1, \beta = 1, \gamma = 0, \delta = 3$

Now,

(A) $\alpha + \beta + \gamma = 1 + 1 + 0 = 2 \ne \delta$

(B) $\alpha + \beta = 1 + 1 = 2$
(C) $\delta + \beta = 3 + 1 = 4$
(D) $\delta - \gamma = 3$