12 DISTINCT balls are to be placed in 3 IDENTICAL boxes, then the probability of one the boxes constrains exactly 3 balls, is ???

Answer is 12C3×2*9/3*12 because total no of choice is 3*12 and no we have to put 12ball in 3 box and one box contain 3 ball means 9 ball are left that should be distributed in 2 box that makes 2*9 . Please approve it