smc

Note that odd sums can only be formed by $(e,e,o)$ or $(o,o,o),$ so we focus on placing the evens: we need to have each even be with another even in each row/column. Because there are only 5 odd numbers overall, we can only have 1 $(o,o,o)$ row and 1 $(o,o,o)$ column. It can be seen that there are $3\cdot 3=9$ ways to do this. There are then $5!$ ways to permute the odd numbers, and $4!$ ways to permute the even numbers, thus giving the answer as $\frac{9\cdot 5!\cdot 4!}{9!}=\boxed{\frac{1}{14}}$.