【Answer】 5 (Each vertex is connected by three edges, so the sum of these three edges is equivalent to adding the number at this vertex three times). Considering $S_{\text {edge }}+S_{\text {vertex }}=1+2+3+4+5+6+7+8+9+11=56$, we can deduce that $S_{\text {vertex }}=14$. Possible combinations are: $1+2+3+8$, $1+2+4+7$, $1+2+5+6$, $1+3+4+6$. Next, we start eliminating:
(1) Since $P+Q=9$, $1+2+5+6$ is eliminated (there are no two numbers that sum to 9).
(2) For $1+2+3+8$, since $1+2=3$, the number on one edge should be 3, but one of the vertex numbers is also 3, which is a contradiction. Similarly, $1+3+4+6$ can be eliminated.

Finally, the only remaining combination is: $1+2+4+7$, leading to the following diagram, so the answer is 5.
12. For a regular tetrahedron $P Q R S$, there are four vertices and six edges, each labeled with a number, making a total of 10 numbers. These 10 numbers are $1, 2, 3, 4, 5, 6, 7, 8, 9, 11$. Each number is used exactly once, and each number on an edge represents the sum of the numbers at the two vertices it connects. The number on edge $P Q$ is 9, then the number on edge $R S$ is $\qquad$ .