jmc

To have an even sum with three numbers, we must add either $E+O+O$, or $E+E+E$, where $O$ represents an odd number, and $E$ represents an even number. Since there are not three even numbers in the given set, $E+E+E$ is impossible. Thus, we must choose two odd numbers, and one even number. There are $2$ choices for the even number. There are $4$ choices for the first odd number. There are $3$ choices for the last odd number. But the order in which we pick these 2 numbers doesn't matter, so this overcounts the pairs of odd numbers by a factor of $2$. Thus, we have $\frac{4\cdot 3}{2}=6$ choices for a pair of odd numbers. In total, there are $2$ choices for an even number, and $6$ choices for the odd numbers, giving a total of $2\cdot 6=12$ possible choices for a 3-element set that has an even sum. This is option $\boxed{12}$.