We have that $\langle\{(1,1,0),(1,-1,0),(0,0,1)\}\rangle = \mathbb{Z}^3$ and, thus, for $s=2$ and $v_2= (1,-1,0), v_1 = (0,0,1)$ and $d_1=6, d_1=1$, we have that $\langle\{d_1v_1,d_2v_2\}\rangle = N$, with $\langle d_1\rangle=6\mathbb{Z} \subset \langle d_2\rangle = 1\mathbb{Z}$.
Inr order to find $M/N$, first of all define $N:A\mathbb{Z}^3$, where $A$ is the matrix
\begin{equation}
A = \left(
\begin{array} {cc}
1& 1 \\
1 &-1 \\
6& 6\nonumber
\end{array}
\right)
\end{equation}
We can use rows and colums operations to find a similar matrix $A'$
\begin{equation}
A'= \left(
\begin{array} {cc}
0& 1 \\
0 &-6 \\
2& 0\nonumber
\end{array}
\right)
\end{equation}
And, clearly, we have that $A = \mathbb{Z} \times 6\mathbb{Z} \times 2\mathbb{Z}$. This way,
\begin{equation}
M/N = \frac{\mathbb{Z}^3}{\mathbb{Z} \times 6\mathbb{Z} \times 2\mathbb{Z}} \simeq \mathbb{Z}_6 \times \mathbb{Z}_2 \simeq \mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}
\nonumber
\end{equation}
and, thus, the invariants are $2\mathbb{Z} = \langle 2 \rangle$ and $6\mathbb{Z} = \langle 6 \rangle$ and the elementary divisors are $2$ and $6$.
In order to find the rank of $(M/N)/T(M/N)$, we note that $T(M/N) = \{(x,y) \in M/N | (x,y).z = 0 \textrm{ for a regular } z \in \mathbb{Z}\}$ which is, in other words, an element $z$ such that $xz \in 6\mathbb{Z}$ and $yz \in 2\mathbb{Z}$, and since for $z \in 6\mathbb{Z}$, every element of $M/N$ would be in $T(M/N)$ and, so, $T(M/N) = M/N$, which implies that $(M/N)/T(M/N) = \{0\}$, whose rank is $1$.
