= Solution
For $a\ne1,5/2$, the <matrix> is invertible, so there is exactly one solution. To exhibit it, subtract the first <linear equation> from the third to obtain $(a-1)x=2$, then subtract twice the second from the first to obtain $-x+(5-2a)z=-3$. Back-substitution gives
$$
\boxed{x=\frac2{a-1},\qquad
z=\frac{3a-5}{(a-1)(2a-5)},\qquad
y=\frac{a^2-10a+15}{(a-1)(2a-5)}}.
$$
At $a=1$, the first and third <linear equations> have identical left-hand sides but right-hand sides one and three, so there is no solution.
At $a=5/2$, subtracting the first from the third forces $x=4/3$, while the first minus twice the second forces $x=3$. These contradict each other, so there is again no solution. Thus there is \b[one solution for $a\ne1,5/2$, no solution for either exceptional value, and no parameter value with infinitely many solutions].
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