Without counting the symmetries, there are 2 ways you could do this. I will use \(\displaystyle{a}+{b}+{c}={13}.\)

The following solutions are to be associated to the pair (a,b,c):

\(\displaystyle{\left({5},{5},{3}\right)},{\left({4},{5},{4}\right)}\) When you count the symmetries, each solution has 6 different arrangements and since there are 2 different solutions, you have a total of 12 solutions (symmetries included).

Here is why:

Since \(\displaystyle{0}≤{a},{b},{c}{<}{6},\) it must be that \(\displaystyle{a}+{b}{<}{12}\) and since no solutions will exist for \(\displaystyle{a}+{b}={11}\), we consider instead that \(\displaystyle{a}+{b}\leq{10}\). From here on, we continue cutting our possibilities to the following:

\(\displaystyle{a}+{b}={10},{c}={3}\)

\(\displaystyle{a}+{b}={9},{c}={4}\)

\(\displaystyle{a}+{b}={8},{c}={5}\)

The rest is just figuring out the values of aa and bb that satisfy each case, which is also easy.