The augmented matrix of this system:
\[
\left(\begin{array}{@{}{cccc}|c@{}}
a-2 & 2 & 3 & 0 & -1\\
0 & 1 & 1 & -1 & 0\\
0 & 0 & 2(1+a) & 1+a & 3
\end{array}\right)
\]
is already in row echelon form. You are correct that if $a=-1$, then the last equation becomes $0z+0w=3$, and it has no solutions. If $a\ne-1$, then we can solve that equation for $z$:
\[
z=\frac{3-(1+a)w}{2(1+a)}
\]
so every value of $w$ gives a unique value of $z$ that satisfies the last equation. From the second equation, given $w$ and $z$, we can find a unique $y$ that satisfies the second equation. Finally, if $a\ne2$, then given $w$, $z$ and $y$ we can find a unique $x$ that satisfies the first equation. All this happens for every value of $w$, so in this case there are infinitely many solutions.
What happens when $a=2$? Consider the system without $x$ (whose augmented matrix lacks the first column) and determine how many solutions it has. But the coefficient of $x$ is 0 in all equations, so $x$ can be arbitrary.
