jmc

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 $12296$, $$\begin{array}{rl}\left(n^2-4n+4\right)+\left(n^2\right)+\left(n^2+4n+4\right)& =12296\\ 3n^2+8& =12296\\ 3n^2& =12288\\ n^2& =4096\\ n& =\pm 64\end{array}$$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 $\boxed{32736}$.