11. Let the four subsets be $M_{i}=\left(a_{i}, b_{i}, c_{i}\right)$, where $a_{i}=b_{i}+c_{i}, b_{i}>c_{i}, i=1,2,3,4$.
Given $a_{1}=27$.
Thus, $10 \leqslant a_{3} \leqslant 11$.
If $a_{3}=10$, then from
$$
a_{1}+a_{2}=17, a_{2}a_{1}+a_{2}=17,
$$

we get $a_{2}=9, a_{1}=8$
$$
\Rightarrow\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(8,9,10,12) .
$$

From $8=b_{1}+c_{1}, 9=b_{2}+c_{2}, 10=b_{3}+c_{3}$, $12=b_{4}+c_{4}$, we must have
$$
b_{4}=11, c_{4}=1 .
$$

This results in two cases:
$$
\begin{array}{l}
12=11+1,10=7+3,9=5+4,8=6+2 ; \\
12=11+1,10=6+4,9=7+2,8=5+3 .
\end{array}
$$

Corresponding to two partition methods:
$$
\begin{array}{l}
(12,11,1),(10,7,3),(9,5,4),(8,6,2) \text {; } \\
(12,11,1),(10,6,4),(9,7,2),(8,5,3) .
\end{array}
$$

If $a_{3}=11$, then $a_{1}+a_{2}=16$.
Thus $9 \leqslant a_{2} \leqslant 10$
$$
\Rightarrow\left(a_{1}, a_{2}\right)=(6,10),(7,9) \text {. }
$$

For $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(6,10,11,12)$, we get three partition methods:
$$
\begin{array}{l}
(12,8,4),(11,9,2),(10,7,3),(6,5,1) ; \\
(12,9,3),(11,7,4),(10,8,2),(6,5,1) ; \\
(12,7,5),(11,8,3),(10,9,1),(6,4,2) .
\end{array}
$$

For $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(7,9,11,12)$, we also get three partition methods:
$$
\begin{array}{l}
(12,8,4),(11,10,1),(9,6,3),(7,5,2) ; \\
(12,10,2),(11,6,5),(9,8,1),(7,4,3) ; \\
(12,10,2),(11,8,3),(9,5,4),(7,6,1) .
\end{array}
$$

Therefore, there are a total of eight partition methods for this problem.
11. Divide the set $M=$ $\{1,2, \cdots, 12\}$ of the first 12 positive integers into four triplets, such that in each triplet, one number is equal to the sum of the other two. Find the number of different ways to do this.