Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

From the diagram,

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

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

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

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

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

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

$\therefore 2\lambda +0=1 \Rightarrow \lambda =\frac{1}{2}$

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

Now,

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

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