= Solution
The equality $x=a^2$ and the relator $[b,x]=1$ show that $x$ commutes with both $a$ and $b$. Since $c=(ab)^{-1}$, it also commutes with $c$, and then with $y=b^3$. Hence $x\in Z(G)$. Similarly, $y=b^3$ commutes with $b$ and, by $[a,y]=1$, with $a$; it therefore commutes with $c$ and $x$. Thus
$$
\langle x,y\rangle\leq Z(G).
$$
Quotienting by $\langle x,y\rangle$ gives
$$
G/\langle x,y\rangle
\cong\langle a,b,c\mid a^2,b^3,c^7,abc\rangle
=\Delta(2,3,7).
$$
A non-elementary <Fuchsian group> has trivial <center of a group>[center]: two hyperbolic elements with different pairs of boundary fixed points have only the identity in their common centralizer in $\operatorname{PSL}_2(\mathbb R)$. Therefore the image in the quotient of every element of $Z(G)$ is trivial. It follows that
$$
Z(G)=\langle x,y\rangle.
$$
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