For which values of $a$ the system has one/none/infinite solutions, write the general solution
$$ \begin{cases} x_1+(a-1)x_2-x_3=4\\ ax_1+(a-1)x_2-x_3=a+3\\ x_1+(a-1)x_2+(a-3)x_3=7 \end{cases} $$
$$ \left( \begin{array}{ccc|c} 1 & a-1 & -1 & 4 \\ 0 & -(a-1)^2 & a-1 & -3a+3 \\ 0 & 0& a-2 & 3 \\ \end{array} \right) $$
So for $a=2$ there is no solution, $a\neq2,1$ there is one solution and for $a=1$ there is infinite solutions in this case we have
$$ \left( \begin{array}{ccc|c} 1 & 0 & -1 & 4 \\ 0 & 0 & -1 & 3 \\ 0 & 0& 0 & 0 \\ \end{array} \right) $$
$$ \begin{cases} x_1-x_3=4\rightarrow x_1=1\\ -x_3=3\rightarrow x_3=-3\\ \end{cases} $$
So the general solution is
$$\begin{pmatrix} 1 \\ t \\ -3\\ \end{pmatrix}=t\begin{pmatrix} 0 \\ 1 \\ 0\\ \end{pmatrix}+\begin{pmatrix} 1 \\ 0\\ -3\\ \end{pmatrix}$$
Is that correct?