Solution
= Solution
Adding three times the second <linear equation> to the third gives $3(3-\alpha)z=0$. Since $\alpha\ne3$, it follows that $z=0$. The second <linear equation> then gives $y=x+1$, and the first becomes $(1+2\alpha)x=1-2\alpha$. Thus the unique solution is
$$
\boxed{x=\frac{1-2\alpha}{1+2\alpha},\qquad y=\frac2{1+2\alpha},\qquad z=0}.
$$
The assumption $\alpha\ne-1/2$ makes the denominator nonzero; $\beta$ is unrestricted.