What are the constraints on $\alpha$ so that $AX=B$ has a solution? I found the following problem and I'm a little confused.

Consider $$A=  \left( \begin{array}{ccc}
3 & 2 & -1 & 5 \\
1 & -1 & 2 & 2\\
0 & 5 & 7 & \alpha \end{array} \right)$$ and $$B=  \left( \begin{array}{ccc}
0 & 3  \\
0 & -1 \\
0 & 6 \end{array} \right)$$
What are the constraints on $\alpha$ so that the matrix equation $AX=B$ has solution?

Since neither $A$ nor $B$ are square, I can't get their inverses. Is the problem wrong?

Ignoring the fourth column, notice that 
$$\begin{pmatrix} 3 & 2 & -1 \\ 1 & -1 & 2\\ 0 & 5 & 7  \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}$$
and
$$\begin{pmatrix} 3 & 2 & -1 \\ 1 & -1 & 2\\ 0 & 5 & 7  \end{pmatrix} \begin{pmatrix} \frac15 \\ \frac65 \\ 0 \end{pmatrix}=\begin{pmatrix}3 \\ -1 \\ 6 \end{pmatrix}.$$
Can you comment on whether $\alpha$ influence existence of solutions? 
Are you able to construct an $X$ for the original problem?
Remark:
In general, you might like to perform row operations on the system of equations. 
Tips:
If $A_1$ is a matrix that consists of some columns of $A$ and $A_1$ is non-singular, the solution to $AX=B$ always exist. Do you see why?
