11. (This sub-question is worth 15 points) Divide the set $M=\{1,2, \cdots, 12\}$ of the first 12 positive integers into four three-element subsets $M_{1}, M_{2}, M_{3}, M_{4}$, such that in each three-element subset, one number is equal to the sum of the other two. Find the number of different ways to do this.
Let the four subsets be $M_{i}=\left(a_{i}, b_{i}, c_{i}\right), i=1,2,3,4$, where $a_{i}=b_{i}+c_{i}, b_{i}>c_{i}, i=1,2,3,4$. Given $a_{1}27$, we have $10 \leq a_{3} \leq 11$. If $a_{3}=10$, then from $a_{1}+a_{2}=17, a_{2}a_{1}+a_{2}=17$, we get $a_{2}=9, a_{1}=8$, which gives $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(8,9,10,12)$. From $8=b_{1}+c_{1}, 9=b_{2}+c_{2}, 10=b_{3}+c_{3}, 12=b_{4}+c_{4}$, we must have $b_{4}=11, c_{4}=1$, resulting in two cases: $12=11+1, 10=7+3, 9=5+4, 8=6+2$ and $12=11+1, 10=6+4, 9=7+2, 8=5+3$, corresponding to two partitions: $(12,11,1),(10,7,3),(9,5,4),(8,6,2)$ and $(12,11,1),(10,6,4),(9,7,2),(8,5,3)$.
(9 points) If $a_{3}=11$, then $a_{1}+a_{2}=16$, so $8<a_{2}<11$, giving $\left(a_{1}, a_{2}\right)=(6,10),(7,9)$. For $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(6,10,11,12)$, we get three partitions: $(12,8,4),(11,9,2),(10,7,3),(6,5,1) ;(12,9,3),(11,7,4),(10,8,2),(6,5,1) ;(12,7,5),(11,8,3),(10,9,1),(6,4,2)$; For $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(7,9,11,12)$, we also get three partitions: $(12,8,4),(11,10,1),(9,6,3),(7,5,2) ;(12,10,2),(11,6,5),(9,8,1),(7,4,3) ;(12,10,2),(11,8,3),(9,5,4),(7,6,1)$; Therefore, there are eight partition schemes in total.
12. (This sub-question is worth 20 points) In $\triangle A B C$, the incenter is $I$, and $D, E, F$ are the midpoints of sides $B C, C A, A B$ respectively. Prove that line $D I$ bisects the perimeter of $\triangle D E F$.

Proof: As shown in the figure, without loss of generality, assume $A B \geq A C$, and the incircle of $\triangle A B C$ touches $B C, C A, A B$ at $T, K_{1}, K_{2}$ respectively. Draw the diameter $T K$ of the incircle, and draw the tangent to $\odot I$ at $K$ intersecting $A C, A B$ at $M, N$ respectively, then $N M \| B C$.

Since $\odot I$ is the excircle of $\triangle A M N$, and $A K_{1}=A K_{2}$, we have $M K=M K_{1}, N K=N K_{2}$, so $A M+M K=A N+N K$.
(5 points)
Extend $A K$ to intersect $B C$ at $G$, then by the properties of the incircle of a triangle: $B G=C T$, thus $D T=D G$, hence $D I$ is the midline of $\triangle T G K$, so $D P \| A G$. Since $B D E F$ is a parallelogram, $\triangle D E P \sim \triangle A B G$, with a similarity ratio of $\frac{D E}{A B}=\frac{1}{2}$.
Similarly, $\triangle D F P \sim \triangle A C G$, with a similarity ratio of $\frac{D F}{A C}=\frac{1}{2}$.
Also note that $\triangle A M K \sim \triangle A C G, \triangle A N K \sim \triangle A B G$, with similarity ratios of $\frac{A K}{A G}$.
(10 points)
Since $A M+M K=A N+N K$, we have $A C+C G=A B+B G$, thus $D F+F P=D E+E P$, which proves the desired result.
(20 points)