= Solution
The same elimination at general $a$ gives the rows
$$
\begin{pmatrix}1&1&1&|&3\\0&0&a-3&|&-2\\
0&1-a&3-a&|&5-3a\end{pmatrix}.
$$
The coefficient <determinant> is $(a-3)(a-1)$. Only $a=1$ or $a=3$ can therefore fail to have a unique solution. At $a=3$, the second row reads $0=-2$, so \b[there is no solution]. At $a=1$, the last two rows are proportional and both specify $z=1$, leaving one free unknown; hence \b[there are infinitely many solutions]. Every other real $a$ gives a unique solution.
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