Three, (50 points) On a plane, there are 12 points, and no three points are collinear. Using any one of these points as the starting point and another as the endpoint, draw all possible vectors. A triangle whose three side vectors sum to the zero vector is called a "zero triangle." Find the maximum number of "zero triangles" that can be formed with these points as vertices.

保留源文本的换行和格式，翻译结果如下：

Three, (50 points) On a plane, there are 12 points, and no three points are collinear. Using any one of these points as the starting point and another as the endpoint, draw all possible vectors. A triangle whose three side vectors sum to the zero vector is called a "zero triangle." Find the maximum number of "zero triangles" that can be formed with these points as vertices.
Three, let these 12 points be $P_{1}, P_{2}, \cdots, P_{12}$. The number of triangles determined by these 12 points is $\mathrm{C}_{12}^{3}$. Let the number of vectors starting from $P_{1}(i=1,2, \cdots, 12)$ be $x_{i}, 0 \leqslant x_{i} \leqslant 11$. If a triangle with 3 points as vertices is a "non-zero triangle", then there is and only one point that is the starting point of the two sides vectors of this triangle. Therefore, the number of "non-zero triangles" with $P_{i}$ as one of the vertices and the starting point of two sides is $\mathrm{C}_{x_{1}^{2}}^{2}$ (with the convention $\mathrm{C}_{1}^{2}=\mathrm{C}_{0}^{2}=0$). Thus, the total number of "non-zero triangles" among the triangles with these points as vertices is $\sum_{i=1}^{12} \mathrm{C}_{x_{i}}^{2}$. Therefore, the number of "zero triangles" is $\mathrm{C}_{12}^{3}-\sum_{i=1}^{12} \mathrm{C}_{x_{i}}^{2}$.

First, find the minimum value of $\sum_{i=1}^{12} C_{x_{i}}^{2}$.
Since $\sum_{i=1}^{12} x_{i}=\mathrm{C}_{12}^{2}=66$, we have
$$
\begin{array}{l}
\sum_{i=1}^{12} \mathrm{C}_{x_{i}}^{2}=\sum_{i=1}^{12} \frac{x_{i}\left(x_{i}-1\right)}{2} \\
=\frac{1}{2} \sum_{i=1}^{12}\left(x_{i}^{2}-x_{i}\right)=\frac{1}{2} \sum_{i=1}^{12} x_{i}^{2}-33 .
\end{array}
$$

Since the non-negative integers $x_{i}$ do not exceed 11, $\sum_{i=1}^{12} x_{i}^{2}$ has a minimum value.
If there exist $1 \leqslant i, j \leqslant 12(i \neq j)$ such that $x_{1}-x_{j} \geqslant 2$, we can denote
$$
x_{i}^{\prime}=x_{1}-1, x_{j}^{\prime}=x_{j}+1 \text {. }
$$

Clearly, $x_{i}>x_{i}^{\prime} \geqslant x_{j}^{\prime}>x_{j}$. Then
$$
\begin{array}{l}
x_{i}^{2}+x_{j}^{2}-\left(x_{1}^{\prime 2}+x_{j}^{\prime 2}\right) \\
=x_{1}^{2}+x_{j}^{2}-\left(x_{i}-1\right)^{2}-\left(x_{j}+1\right)^{2} \\
=2\left(x_{i}-x_{j}-1\right)>0 .
\end{array}
$$

Since $x_{i} \in \mathbf{N}$, for all $1 \leqslant i, j \leqslant 12$, $\sum_{i=1}^{12} x_{i}^{2}$ takes its minimum value only when $\left|x_{i}-x_{j}\right|=0$ or 1, i.e., when
$$
\left\{x_{1}, x_{2}, \cdots, x_{12}\right\}=\{5,5,5,5,5,5,6,6,6,6,6,6\}
$$

$\sum_{i=1}^{12} x_{i}^{2}$ takes its minimum value $5^{2} \times 6+6^{2} \times 6=366$.
Therefore, the minimum value of $\sum_{i=1}^{12} \mathrm{C}_{x_{i}}^{2}$ is $\frac{1}{2} \times 366-33=150$.
Thus, the maximum number of "zero triangles" is
$$
\mathrm{C}_{12}^{3}-150=70 \text {. }
$$

Note: In this problem, since $x_{i} \in \mathbf{N}$, the minimum value of $\sum_{i=1}^{12} x_{i}^{2}$ cannot be found using the mean inequality. Therefore, this minimum value is not $\frac{66^{2}}{12}=363$.