9. (10 points) On the blackboard, there are 11 ones, 22 twos, 33 threes, and 44 fours. Perform the following operation: each time, erase 3 different numbers, and write 2 more of the fourth number that was not erased. For example: if in one operation, 1 one, 1 two, and 1 three are erased, then write 2 more fours. After several operations, only 3 numbers remain on the blackboard, and no further operations can be performed. What is the product of the last three remaining numbers? $\qquad$ .
【Analysis】This problem belongs to an operational issue. The simplest case to work backwards is when there are only 3 numbers left on the blackboard, and no further operations can be performed according to the rules. There are only two scenarios: 2 $a$s and 1 $b$, or 3 $a$s. We perform operations based on the rules, and when only 3 numbers remain and no further operations can be performed, the problem is solved.

【Solution】Solution: The reasoning is listed as follows:
\begin{tabular}{|r|r|r|r|}
\hline 1 & 2 & 3 & 4 \\
\hline 11 & 22 & 33 & 44 \\
\hline 33 & 11 & 22 & 33 \\
\hline 22 & 33 & 11 & 22 \\
\hline 11 & 22 & 33 & 11 \\
\hline 0 & 11 & 22 & 33 \\
\hline 22 & 0 & 11 & 22 \\
\hline 11 & 22 & 0 & 11 \\
\hline 5 & 16 & 12 & 5 \\
\hline 0 & 11 & 7 & 15 \\
\hline 14 & 4 & 0 & 8 \\
\hline 10 & 0 & 8 & 4 \\
\hline 6 & 8 & 4 & 0 \\
\hline 2 & 4 & 0 & 8 \\
\hline
\end{tabular}
\begin{tabular}{|l|l|l|l|}
\hline 0 & 2 & 4 & 6 \\
\hline 4 & 0 & 2 & 4 \\
\hline 2 & 4 & 0 & 2 \\
\hline 1 & 3 & 2 & 1 \\
\hline 3 & 2 & 1 & 0 \\
\hline 2 & 1 & 0 & 2 \\
\hline 1 & 0 & 2 & 1 \\
\hline 0 & 2 & 1 & 0 \\
\hline
\end{tabular}

In the end, there are 2 2s and 1 3 left, and their product is:
$$
2 \times 2 \times 3=12 ;
$$

Therefore, the answer is: 12.