Consider the equation $(a-3)x=a^2-9$, where a is a constant. Which statement is correct?

Since $a^2-9=(a-3)(a+3)$, the equation is $(a-3)x=(a-3)(a+3)$. If a ≠ 3 the unique solution is $x=a+3$. If $a=3$ it becomes 0 · $x=0$, which every real $x$ satisfies — infinitely many solutions. (At $a=-3$ it reads $-6x=0$, so $x=0$ is a genuine unique solution.)