= Solution
{c}
Writing the unknown <vector> as $(x,y,z)^T$, the first two <linear equations> give $x=z-1$ and $y=z-\beta$. Substitution into the third gives
$$
-(z-1)-(z-\beta)+\alpha z=1,\qquad(\alpha-2)z=-\beta.
$$
Thus, when $\alpha\ne2$, put $t=-\beta/(\alpha-2)$. There is exactly one solution:
$$
\boxed{(x,y,z)=(t-1,t-\beta,t),\qquad t=-\frac{\beta}{\alpha-2}.}
$$
When $\alpha=2$, the remaining equation is $0=-\beta$. Hence
$$
\boxed{\begin{aligned}
\alpha=2,\ \beta\ne0&:\quad\varnothing,\\
\alpha=2,\ \beta=0&:\quad\{(-1,0,0)+t(1,1,1):t\in\mathbb R\}.
\end{aligned}}
$$
The second case is an <affine line in a vector space>, so it contains infinitely many solutions. These cases exhaust all <real numbers> $\alpha,\beta$.
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