Determine a direct product of cyclic groups that is isomorphic to $H$.
The Smith Normal form can be used to find the invariant factors in the structure theorem for finitely generated abelian groups. I set up the following matrix $$\begin{pmatrix} 24 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 3 \\ 10 & 3 & 2\end{pmatrix}$$ but I was unable to find the SMF of this matrix.
I know the following about the structure of $H$:
$|\langle (10,3,2)\rangle|=12$
Also $|G|=24*6*3$ so by Lagrange's Theorem, $|H| = 36$.
By the fundamental theorem of finitely generated abelain groups, $H$ may have the following structure:
$\mathbb{Z}/36\mathbb{Z}$
$\mathbb{Z}/18\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$
$\mathbb{Z}/12\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$
$\mathbb{Z}/6\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}$
I am not sure how to identify the correct group structure without the invariant factors from the SMNF.
Edit: Here is some of my work by hand. Am I proceeding correctly?
\begin{align*} &\, \begin{pmatrix} 24 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 3 \\ 10 & 3 & 2 \end{pmatrix} \sim \begin{pmatrix} 10 & 3 & 2 \\ 24 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 3 \end{pmatrix} \sim \begin{pmatrix} 2 & 10 & 3 \\ 0 & 24 & 0 \\ 0 & 0 & 6 \\ 3 & 0 & 0 \end{pmatrix} \\ \sim&\, \begin{pmatrix} 2 & 10 & 1 \\ 0 & 24 & 0 \\ 0 & 0 & 0 \\ 3 & 0 & -3 \end{pmatrix} \sim \begin{pmatrix} 1 & 2 & 10 \\ 0 & 0 & 24 \\ 0 & 0 & 0 \\ -3 & 3 & 0 \end{pmatrix} \sim \begin{pmatrix} 1 & 2 & 10 \\ 0 & 0 & 24 \\ 0 & 0 & 0 \\ 0 & 9 & 30 \end{pmatrix} \\ \sim&\, \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 24 \\ 0 & 0 & 0 \\ 0 & 9 & 30 \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 0 \\ 0 & 9 & 30 \\ 0 & 0 & 24 \\ 0 & 0 & 0 \end{pmatrix} \end{align*}
(Original image here.)
The Smith Form has diagonal entries 1, 3, 12 so that the quotient is isomorphic to $Z_3\times Z_{12}$.