Three. (25 points) As shown in Figure 5, there are 12 equally spaced points on the circumference of a clock face, marked with the numbers $1, 2, \cdots, 12$ in sequence. Please use these equally spaced points as vertices to form 4 triangles (dividing these 12 equally spaced points into 4 groups) such that the following conditions are met:
(1) No two triangles share a common vertex;
(2) In each triangle, one of the numbers marked on the vertices is equal to the sum of the other two numbers.
Find all different grouping schemes.
(A triangle with vertices $a, b, c$ can be represented as $(a, b, c)$.)
Three, let 4 triangles be $\left(a_{i}, b_{i}, c_{i}\right), i=1,2, 3,4$, where $a_{i}=b_{i}+c_{i}, b_{i}>c_{i}$.
Let $a_{1}=27$.
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 $a_{2}=9, a_{1}=8$, that is
$$
\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(8,9,10,12) \text {. }
$$

From $8=b_{1}+c_{1}, 9=b_{2}+c_{2}, 10=b_{3}+c_{3}, 12=b_{4}+c_{4}$, it must be that $b_{4}=11, c_{4}=1$, yielding 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 divisions:
$$
\begin{array}{l}
(12,11,1),(10,7,3),(9,5,4),(8,6,2) ; \\
(12,11,1),(10,6,4),(9,7,2),(8,5,3) .
\end{array}
$$

If $a_{3}=11$, then $a_{1}+a_{2}=16$.
Thus, $8<a_{2}<11$, yielding respectively
$$
\left(a_{1}, a_{2}\right)=(6,10),(7,9) \text {. }
$$

For $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(6,10,11,12)$, we get three divisions:
$(12,8,4),(11,9,2),(10,7,3),(6,5,1)$; $(12,9,3),(11,7,4),(10,8,2),(6,5,1)$; $(12,7,5),(11,8,3),(10,9,1),(6,4,2)$.
For $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(7,9,11,12)$, we also get three divisions:
$(12,8,4),(11,10,1),(9,6,3),(7,5,2)$; $(12,10,2),(11,6,5),(9,8,1),(7,4,3)$; $(12,10,2),(11,8,3),(9,5,4),(7,6,1)$.
Therefore, there are eight grouping schemes in total.
(Tao Pingsheng, Department of Mathematics and Computer Science, Jiangxi Science and Technology Normal University, 330013)