Given two of the numbers, only certain choices for the third number will give you a multiple of 3 as your total. It appears to me that regardless of what the first two numbers total to, roughly one third of the remaining numbers will be suitable.
There are thus 11 choices for the first number, 10 choices for the second number, and roughly 3 valid choices for the third number, yielding about 330 possible sequences.
If you're saying that the ordering doesn't matter, then I'd guess you need to divide 330 by the number of possible ways that 3 items can be rearranged (i.e., 6). That gives 55.
That's my guess. Now to go try an exhaustive search and see what the actual answer is...
Edit: It appears the exact answer is 342 ordered combinations, or 57 unordered ones. This is close to my estimate above.