= 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 $x^my^nz^k=1$ exactly when
$$
(m,n,k)=a(2,6,0)+b(1,2,2)\qquad(a,b\in\mathbb Z).
$$
Eliminating $a,b$ 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. Therefore
$$
\boxed{x^my^nz^k=1\ \Longleftrightarrow\ n\equiv0\pmod2\quad\text{and}\ k+2n-6m=0.}
$$
One can also see the invariant factors directly. The second relation gives $x=y^{-2}z^{-2}$. Substitution into the first gives $y^2z^{-4}=1$. Put $t=yz^{-2}$; then
$$
\boxed{G=\langle t,z\mid t^2=1,\ [t,z]=1\rangle\cong C_2\times\mathbb Z,\qquad
x=z^{-6},\quad y=tz^2.}
$$
The 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 $(n\bmod2,k+2n-6m)$, giving a terminating <word problem for a group> algorithm.
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