$\Delta_{1}=\left|\begin{array}{lll}2 & 1 & \alpha \\ 4 & 1 & 1 \\ 1 & 0 & 2\end{array}\right|=-(3+\alpha)$

$\Delta_{2}=\left|\begin{array}{lll}1 & 2 & \alpha \\ 3 & 4 & 1 \\ 1 & 1 & 2\end{array}\right|=-(\alpha+3)$

$\Delta_{3}=\left|\begin{array}{lll}1 & 1 & 2 \\ 3 & 1 & 4 \\ 1 & 0 & 1\end{array}\right|=0$

$\alpha \neq-3, x=1, y=1, z=0$,

Now points $(\alpha, 1),(1, \alpha) \ and (1,-1)$ are collinear

$\left|\begin{array}{lcc}\alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & -1 & 1\end{array}\right|=0$

$\Rightarrow \alpha(\alpha+1)-1(1-1)+1(-1-\alpha)=0$

$\alpha^{2}+\alpha-1-\alpha=0$