find a value of $a$ so the equation below has some solutions. I need find a value of $a$ so the equation below has some solutions.
$\begin{pmatrix}
3 & 1 &-5&2\\
-1 & 0 &2&-1\\
2&1&-3&1\\
3&-2&4&-1
\end{pmatrix}
\left(
  \begin{array}{c}
    x_1 \\
    x_2 \\
    x_3 \\
    x_4 \\
    \end{array}
\right)=
\left(
  \begin{array}{c}
    2a-3 \\
    -a+1 \\
    a-2 \\
    -2a+5 \\
    \end{array}
\right)$
I have tried that find the value by setting$$\operatorname{rank}(A)=\operatorname{rank}(\tilde A)$$
where $\tilde A$ is an enlarged coefficient matrix of $A$.
Then I have found $\operatorname{rank}(A)=3$ by row manipulation.
Similarly, I did the row manipulation and ended up with
$$\begin{pmatrix}
1 & 0 &0&1&-a-3\\
0 & 1 &0&0&1\\
0&0&1&-1&-a-1\\
0&0&0&0&-2
\end{pmatrix}$$
Doesn't this mean that we cannot have solution whatever the value is as we end up with $0=2$ at the bottom line?
Can anyone help me?

You would be right, but you made a mistake in your computations. Last row should be all zeroes. Actually, the RREF is
$$
\begin{bmatrix}
1&0&0&0&-a/6+1/3\\
0&1&0&-1/2&-5a/12-2/3\\
0&0&1&-1/2&-7a/12+2/3\\
0&0&0&0&0
\end{bmatrix}.
$$
