$9 、$ Place 12 small balls labeled with the 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 labeled with the numbers $6, 11$, B has two balls labeled with the numbers $4, 8$, and C has one ball labeled with the number 1. Therefore, the numbers on the other three balls that C has are ( ).

【Analysis】The sum of the numbers on the 4 balls each person takes is $(1)+2+3+\cdots+12 \div 3=26$. Jia already has 17, and still needs 9, which can be chosen from (1, 8) (2, 7) (3, 6) (4, 5). However, 1, 4, 6, and 8 have already been taken, so 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). However, 2, 4, 6, 8, and 11 have already been taken, so Yi can only choose (5, 9). The remaining numbers 3, 10, and 12 are the numbers on the other three balls for Bing.