The plane can be written as $\vec{r}=\hat{k}+t(-\hat{i}+\hat{j})+p(-\hat{i}+\hat{k})$, with a point $\hat{k}$ and direction vectors $(-1,1,0)$ and $(-1,0,1)$.

Its normal is $(-\hat{i}+\hat{j})\times (-\hat{i}+\hat{k})=\hat{i}+\hat{j}+\hat{k}$, so the plane equation is $x+y+z=1$.

For reflection of $Q=(10,15,20)$ in this plane, using the standard formula with normal direction ratios $(1,1,1)$:

$\frac{\alpha -10}{1}=\frac{\beta -15}{1}=\frac{\gamma -20}{1}=-\frac{2(10+15+20-1)}{1^2+1^2+1^2}=-\frac{2\cdot 44}{3}=-\frac{88}{3}.$

Hence $\alpha =10-\frac{88}{3}=-\frac{58}{3}$, $\beta =15-\frac{88}{3}=-\frac{43}{3}$, $\gamma =20-\frac{88}{3}=-\frac{28}{3}$.

Verify each option:

$3(\alpha +\beta )=-58-43=-101$ $✓$ (A is TRUE).

$3(\beta +\gamma )=-43-28=-71$ $✓$ (B is TRUE).

$3(\gamma +\alpha )=-28-58=-86$ $✓$ (C is TRUE).

$3(\alpha +\beta +\gamma )=-58-43-28=-129\ne -121$ (D is FALSE).

$\therefore$ The correct options are (A), (B) and (C).