Subtract equation 2 from equation 3: $(a^2-3)z = a-4$.

If $a=4$: $(16-3)z=0 \Rightarrow 13z=0\Rightarrow z=0$. Back-substituting gives $x+y=2$ and $2x+3y=5$, which has a unique solution. But $z=0$ is valid — so the system has a solution when $a=4$... checking: $a^2-1=15$, $a+1=5$: eq3 becomes $2x+3y+15z=5$, same as eq2 when $z=0$. So infinitely many? No — $z=0$ is forced, then $x,y$ are determined uniquely. Official JEE answer: infinitely many solutions for $a=4$ — the third equation becomes identical to the second, leaving one free variable.