= Solution
The <group presentation> gives the <quotient group> $\mathbb Z^3$ by the subgroup generated by the rows of
$$
M=\begin{pmatrix}3&6&3\\0&9&9\\-3&3&6\end{pmatrix}.
$$
For the <Smith normal form>, the first <determinantal divisor> is the greatest common divisor of all entries, namely $3$. The nonzero two-by-two <matrix minors> are $\pm27$, so the second <determinantal divisor> is $27$. The third row is the second minus the first, and a nonzero minor shows the <rank> is exactly two. Consequently the nonzero invariant factors are $3$ and $27/3=9$, with one free generator. The <Fundamental theorem of finitely generated abelian groups> gives
$$
\boxed{G\cong\mathbb Z\times\mathbb Z/3\mathbb Z\times\mathbb Z/9\mathbb Z.}
$$
An element killed by $3$ has zero free coordinate, arbitrary coordinate in $\mathbb Z/3\mathbb Z$, and coordinate $0,3$ or $6$ in $\mathbb Z/9\mathbb Z$. There are nine such elements. Removing the identity leaves \b[eight elements of order three].
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