You can choose $3$ out of $12$ balls for the first box, $3$ out of $9$ for the second...and so on...But since the boxes are similar you should divide the result by all the possible combinations of the $4$ boxes ($4!$), so the number requested is: $$\frac{\binom{12}{3}\cdot\binom{9}{3}\cdot\binom{6}{3}\cdot\binom{3}{3}}{4!}=\frac{\frac{12!}{9!\cdot 3!}\cdot\frac{9!}{6!\cdot3!}\cdot\frac{6!}{3!\cdot3!}\cdot\frac{3!}{3!\cdot0!}}{4!}=\frac{12!}{3!^4\cdot4!}=15400$$ If the boxes weren't similar the solution would be: $$\binom{12}{3}\cdot\binom{9}{3}\cdot\binom{6}{3}\cdot\binom{3}{3}=\frac{12!}{9!\cdot 3!}\cdot\frac{9!}{6!\cdot3!}\cdot\frac{6!}{3!\cdot3!}\cdot\frac{3!}{3!\cdot0!}=\frac{12!}{3!}=369600$$