Let$$ A = \begin{pmatrix} 2 & -1 & -13 \\ 5 & -2 & -31 \\ -2 & 4 & 22 \\ \end{pmatrix} $$
For which values of $$b = \begin{pmatrix} b_1\\ b_2 \\ b_3 \\ \end{pmatrix} $$ is $Ax = b$ consistent ? I tried to set the matrix in augmented form $$ \left[ \begin{array}{ccc|c} 2&-1&-13&b1\\ 5&-2&-31&b2\\ -2&4&22&b3 \end{array} \right] $$ and then reduce it to get $$ \left[ \begin{array}{ccc|ccc} 1&0&-25/2&(21b1)/2-4b2\\ 0&1&3&-5b1+2b2\\ 0&0&0&16b1-6b2+b3 \end{array} \right] $$ I know that 16b1-b2+b3 = 0 for Ax=b to be consistent, but the question seems to be asking about the values of b1,b2,b3 ?
Since $Ax$ is the linear combination of the columns of $A$ with the coefficients of $x$, one can have $Ax=b$ precisely when $b$ is some linear combination of those columns, in other words when it is in their span. That is already a correct answer, but you can try to describe that span more precisely; three vectors in $\Bbb R^3$ often span the whole space so if that were the case, you could just say "always". But these columns happen not to, so you could try to find either a basis or (better) an equation describing the span. (An equation is better, because you can immediately apply it to a candidate en tell without further ado whether the system is consistent for it.)