 Solution

The same elimination at general $a$ gives the rowsThe 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 there is no solution. At $a=1$, the last two rows are proportional and both specify $z=1$, leaving one free unknown; hence there are infinitely many solutions. Every other real $a$ gives a unique solution.