Equation of plane passing through
$A(0,0,2), B(1,0,1)$ and $C(3,1,1)$ is given by
$\begin{aligned} & \left|\begin{array}{lll}x-0 & y-0 & z-2 \\ 1-0 & 0-0 & 1-2 \\ 3-0 & 1-0 & 1-2\end{array}\right|=0 \\ & \Rightarrow \quad\left|\begin{array}{ccc}x & y & z-2 \\ 1 & 0 & -1 \\ 3 & 1 & -1\end{array}\right|=0 \\ & \Rightarrow \quad x-2 y+z-2=0 \\ & \end{aligned}$
Now cquation of $X Y$ plane is $z=0$
$\begin{aligned} & \therefore \quad \cos \alpha=\frac{0+0+1}{\sqrt{1+4+1} \sqrt{0+0+1}} \\ & \Rightarrow \quad \cos \alpha=\frac{1}{\sqrt{6}}\end{aligned}$
and cquation of $X Z$ plane is $y=0$
$\therefore \quad \cos \beta=\frac{0-2+0}{\sqrt{1+4+1} \sqrt{0+1+0}}=\frac{-2}{\sqrt{6}}$
$\therefore \sin ^2 \alpha+\sin ^2 \beta=2-\cos ^2 \alpha-\cos ^2 \beta$
$=2-\frac{1}{6}-\frac{4}{6}=2-\frac{5}{6}=\frac{7}{6}$