 Solution

All three generators commute, so the group is the abelian group ${\mathbb{Z}}^3$ modulo the integer relation lattice generated by $(2,6,0)$ and $(1,2,2)$. Thus $x^m{y}^n{z}^k=1$ exactly whenEliminating $a,b$ gives $6m-2n-k=0$, and $n=6a+2b$ is even. Conversely, if $n$ is even and $k=6m-2n$, take $a=n/2-m$ and $b=3m-n$; these are integers and reproduce all three coordinates. ThereforeOne can also see the invariant factors directly. The second relation gives $x=y^{-2}{z}^{-2}$. Substitution into the first gives $y^2{z}^{-4}=1$. Put $t=y{z}^{-2}$; thenThe two presentations are equivalent by the explicit substitutions in both directions, so this is not just a quotient guess. An arbitrary input word is first converted, using commutativity, to its exponent totals $m,n,k$. Its coordinates are $(n\mathrm{mod}2,k+2n-6m)$, giving a terminating word problem for a group algorithm.