Solution
= Solution
Apply the determinant-preserving <row operation> $R_3\leftarrow R_3+3R_2$. The new third row is $(0,0,9-3\alpha)$, so expansion of the <determinant> along it gives
$$
\det A=(9-3\alpha)\det\begin{pmatrix}1&2\alpha\\1&-1\end{pmatrix}
=\boxed{3(\alpha-3)(2\alpha+1)}.
$$
Therefore
$$
\boxed{\det A=0\iff\alpha=3\text{ or }\alpha=-\frac12,\qquad\beta\text{ arbitrary}}.
$$
The parameter $\beta$ does not affect the <determinant>.