Alternate approach using subgroup maximality and the product formula.
Since $[G:H]=3$ is prime, $H$ is a maximal subgroup of $G$: any subgroup strictly containing $H$ would have index dividing 3, and the only divisor of 3 other than 3 itself is 1, so that subgroup would have to be all of $G$. The same holds for $K$.
Because $G$ is abelian, $HK=\{hk : h\in H, k\in K\}$ is automatically a subgroup of $G$. Since $H\ne K$ and both have the same order, $K\not\subseteq H$ (otherwise $K=H$). So $HK$ strictly contains $H$. By maximality of $H$, this forces
\[HK=G.\]
Now apply the standard counting formula for the product of two subgroups:
\[|HK|=\frac{|H|\,|K|}{|H\cap K|}.\]
Let $|G|=n$. Since $[G:H]=[G:K]=3$, we have $|H|=|K|=n/3$. Substituting $|HK|=|G|=n$:
\[n=\frac{(n/3)(n/3)}{|H\cap K|} \implies |H\cap K|=\frac{n^2/9}{n}=\frac{n}{9}.\]
Therefore the index is
\[[G:H\cap K]=\frac{|G|}{|H\cap K|}=\frac{n}{n/9}=9.\]
\[\boxed{[G:H\cap K]=9}\]