= Solution
The last equation gives $x=1/2-y$, and the first then gives $y+3z=3/2$. Together with the middle equation this reduces the consistency question to $(1-3\alpha)z=\beta-3\alpha/2$.
If $\alpha\ne1/3$, the <linear system> has the unique solution
$$
\boxed{(x,y,z)=\frac{(6\beta-3\alpha-2,\ 3-6\beta,\ 2\beta-3\alpha)}{2(1-3\alpha)}.}
$$
If $\alpha=1/3$ and $\beta=1/2$, the middle equation repeats one third of $y+3z=3/2$, giving the <affine line> of solutions
$$
\boxed{(x,y,z)=(-1+3s,\ 3/2-3s,\ s),\qquad s\in\mathbb R.}
$$
If $\alpha=1/3$ and $\beta\ne1/2$, the two equations contradict each other and \b[there is no solution]. These are three possible types of solution set, rather than three individual solutions for one fixed choice of parameters.
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