Solve the matrix. The given matrix is matrix $Q$. I performed RREF and got $C$.
Would $C$ yield a unique solution, or are there more steps to the problem?
$$ Q = \begin{pmatrix} 1&1&1&-2&3\\ 2&1&3&2&9\\ 0&-1&1&6&3 \end{pmatrix} $$
$$ C = \begin{pmatrix} 1&0&2&4&6\\ 0&1&-1&-6&-3\\ 0&0&0&0&0 \end{pmatrix} $$
I'm going to assume that $Q$ represents the system \begin{align} x_1 + x_2 + x_3 - 2x_4 &= 3 \\ 2x_1 + x_2 + 3x_3 + 2x_4 &= 9\\ -x_2 + x_3 + 6x_4 &= 3 \end{align} as it's common practice in most undergrad level linear algebra texts to adjoin the RHS to the right of the coefficient matrix for solving a system by Gaussian elimination (i.e. row reduction).
The RREF being $C$ tells us that \begin{align} x_1 + 2x_3 + 4x_4 &= 6 \\ x_2 - x_3 - 6x_4 &= -3 \end{align}
We can solve for the leading variable in each equation to obtain \begin{align} x_1 &= 6 - 2x_3 - 4x_4 \\ x_2 &= -3 + x_3 + 6x_4 \end{align}
Note that you have two free variables, $x_3$ and $x_4$. You can choose any value for each of them and have a valid solution. Thus, rather than a unique solution you have infinitely many.
If I've misinterpreted $Q$ and your original RHS was a zero vector, then the specifics are a little different (three free variables rather than two, and the solved system would look a bit different), but you'd still have infinitely many solutions.