1. You have two atoms of Cr on both sides of the oxidation equation - but you have lost 2 on the RHS of the full equation.

2. No idea why do you think it should be 6 INSTEAD of 3. 3 is wrong - your reduction reaction after being multiplied by 3 will give you 6 atoms of Br (more precisely - six Br^{-} anions) on the RHS.

Seems like you did the same mistake twice.