weigh R1+W1 against R2+B2.
if they balance, then we have either R:1 W:2 B:2 (meaning red ball 1 is the heavy one, etc.) or R:2 W:1 B:1. Weighing pretty much anything against anything else will distinguish these two.
On the other hand,
if one side is heavier, WLOG it's R1+W1 > R2+B2. Then we necessarily have R:1, and the others can't be W:2 B:2. So we have one of W:1 B:1, W:1 B:2, W:2 B:1. Now weigh W1 against B1. In those three cases we have respectively W1 = B1, W1 > B1, W1 < B1.
no third weighing is necessary and this paragraph is just here so it isn't too obvious from the structure of my answer what the desired number is
I suppose I should say explicitly that if the first weighing yields R1+W1 < R2+B2 instead, we just need to swap "white" and "blue" and "light" and "heavy" in the second-weighing strategy above
we are done, in all cases, after at most two weighings.
the first weighing here is essentially the only possibility that makes two weighings enough.