$\begin{aligned} & P=\frac{|2-1-4|}{\sqrt{1^2+1^2}}=\frac{3}{\sqrt{2}} \\ & \sin \left(\frac{\pi}{6}\right)=\frac{3 / \sqrt{2}}{A B}=\frac{1}{2} \Rightarrow A B=\frac{6}{\sqrt{2}} \\ & \Rightarrow(\alpha-1)^2+(\alpha+4-2)^2=18 \end{aligned}$ $\Rightarrow 2 \alpha^2+2 \alpha-13=0 \rightarrow \alpha$ and $\gamma$ satisfy same equation $\begin{aligned} & \Rightarrow \alpha^2+\gamma^2=(\alpha+\gamma)^2-2 \alpha \gamma \\ & =(-1)^2-2\left(\frac{-13}{2}\right)=1+13=14 \end{aligned}$