9. Put 12 small balls marked with numbers $1, 2, 3, \cdots \cdots, 12$ into a paper box. Three people, A, B, and C, each take 4 balls from the box. It is known that the sum of the numbers on the balls they each took is equal. A has two balls marked with numbers 6 and 11, B has two balls marked with numbers $4$ and $8$, and C has one ball marked with number 1. Therefore, the numbers on the other three balls C has are ( ).
From (1, 8) (2, 7) (3, 6) (4, 5), choose 1 group, and since 1, 4, 6, 8 have already been taken, Jia can only choose $(2, 7)$. Yi already has 12, and still needs 14, which can be chosen from (2, 12) (3, 11) (4, 10) (5, 9) (6, 8), and since $2, 4, 6, 8, 11$ have already been taken, Yi can only choose (5, 9). The remaining numbers 3, 10, 12 are on the other three balls for Bing.