Row Reduction & Rank · Hard problem

A $3\times3$ system $A\mathbf x=\mathbf b$ has $A=\begin{pmatrix}2&1&0\\1&2&1\\0&1&2\end{pmatrix}$. For $\mathbf b_1=(5,6,5)$, the solution is $\mathbf x_1=(2,1,2)$. A second right-hand side $\mathbf b_2=(5,6,1)$ agrees with $\mathbf b_1$ in its first two entries and differs only in the third. Find the second component of the solution $\mathbf x_2$ to $A\mathbf x_2=\mathbf b_2$.