9. There are 11 ones, 22 twos, 33 threes, and 44 fours on the blackboard. 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 you erase 1 one, 1 two, and 1 three, 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: Operation problem.
By backtracking the simplest case, there are only 3 numbers left on the blackboard, and it is impossible to continue the operation according to the rules. There are only two situations: 2 $\mathrm{a}$s and one b, or 3 $\mathrm{a}$s. We perform the operation according to the rules. When only 3 numbers are left and cannot continue the operation, the problem is solved.
\begin{tabular}{|l|l|l|l|}
\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
\end{tabular}
\begin{tabular}{|l|l|l|l|}
\hline 1 & 2 & 3 & 4 \\
\hline 6 & 8 & 4 & 0 \\
\hline 2 & 4 & 0 & 8 \\
\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}

So the answer is $2 \times 2 \times 3=12$