Find constants in system of equations

The system can be written as$$\begin{pmatrix} 2& 1 &-1& 2\\ 3&1 &-2& 1\\ 1&1& 3&-1\end{pmatrix} \begin{pmatrix} x\\ y \\ z \\ s\end{pmatrix} = \begin{pmatrix} q_1\\ q_2 \\ q_3\end{pmatrix}.$$ Now since we have that $$\mathrm{rank}\begin{pmatrix} 2& 1 &-1& 2\\ 3&1 &-2& 1\\ 1&1& 3&-1\end{pmatrix} =3$$ the system must be consistent, independently of the values of $q_1,q_2,q_3$. (Obviously it is $$\mathrm{rank}\begin{pmatrix} 2& 1 &-1& 2\\ 3&1 &-2& 1\\ 1&1& 3&-1\end{pmatrix} =\mathrm{rank}\begin{pmatrix} 2& 1 &-1& 2 &q_1\\ 3&1 &-2& 1 & q_2\\ 1&1& 3&-1 & q_3\end{pmatrix}=3.)$$

EDIT

$$\begin{cases} 2x&+y&-z&+2s &= q_1\\ 3x&+y&-2z&+s &= q_2\\ x&+y&+3z&-s &= q_3 \end{cases} \iff \begin{cases} 2x&+y&-z &= q_1-2s\\ 3x&+y&-2z&= q_2-s\\ x&+y&+3z&= q_3+s \end{cases} $$

$$\iff \begin{cases} 2x&+y&-z &= q_1-2s\\ x&&-z&= q_2-q_1+s\\ -x&&+4z&= q_3-q_1+3s \end{cases}$$

$$\iff \begin{cases} 2x&+y&-z &= q_1-2s\\ x&&-z&= q_2-q_1+s\\ &&3z&= q_3+q_2-2q_1+4s \end{cases}$$

From the last equation we get

$$z=\dfrac{q_3+q_2-2q_1+4s}{3}.$$

Now $$x=z+q_2-q_1+s=\dfrac{q_3+4q_2-5q_1+7s}{3}.$$

Finally we get $$y=z-2x+q_1-2s=\dfrac{-q_3-7q_2+11q_1-16s}{3}.$$ That is, the system is consistent independently of the values of $q_1,q_2,q_3.$