R is mid point of $P Q$
$\therefore \mathrm{R}(2,1,-1)$ and it lies on plane equation of plane is $\alpha a+\beta y+\gamma z=\delta$
$ \begin{equation*} \therefore 2 \alpha+\beta-\gamma=\delta \tag{1} \end{equation*} $
Normal vector to plane is
$ \begin{align*} & \overrightarrow{\mathrm{n}}=2 \mathrm{i}+2 \mathrm{j} \\ & \therefore \frac{\alpha}{2}=\frac{\beta}{2}=\frac{\gamma}{0}=\mathrm{k} \\ & \therefore \alpha=2 \mathrm{k}, \beta=2 \mathrm{k}, \gamma=0 \tag{2} \end{align*} $
and $\alpha+\gamma=1$ (given)
from (2) and (3)
$ \therefore \quad \alpha=1, \beta=1, \gamma=0 $
and from (1)
$ \begin{aligned} & 2(1)+1-0=\delta \\ & \delta=3 \end{aligned} $
Now :
$ \begin{aligned} & \alpha+\beta=2 \\ & \delta-\gamma=3 \\ & \delta+\beta=4 \end{aligned} $
so, $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are correct.