The sum of the squares of three consecutive positive even numbers is $12296$. Find the product of the three numbers divided by $8$.
If $n$ is the middle number of the three, the other two numbers are $n-2$ and $n+2$. Therefore, the squares are $n^2-4n+4$, $n^2$, and $n^2+4n+4$. Setting the sum of the three squares equal to $12296$, \begin{align*}
\left(n^2-4n+4\right)+\left(n^2\right)+\left(n^2+4n+4\right)&=12296\\
3n^2+8&=12296\\
3n^2&=12288\\
n^2&=4096\\
n&=\pm64
\end{align*}Because $n$ is positive, $n$ must be $64$. Therefore, the set of numbers is $62, 64, 66$. The product of those is $261888$. The product divided by 8 is $\boxed{32736}$.