Three, (50 points) On a plane, there are 12 points, no three of which are collinear. Taking any one of these points as the starting point and another as the endpoint, construct all such vectors. If the sum of the three side vectors of a triangle is the zero vector, then the triangle is called a "zero triangle." Find the maximum number of zero triangles that can be formed with these 12 points as vertices.
Three, let these 12 points be $P_{1}, P_{2}, \cdots, P_{12}$, the number of triangles determined by them is $\mathrm{C}_{12}^{3}$.

Let the number of vectors starting from $P_{i}(i=1,2, \cdots, 12)$ be $x_{i}\left(0 \leqslant x_{i} \leqslant 11\right)$.

If a triangle with three points as vertices is a non-zero triangle, then exactly one point 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 $\mathrm{C}_{x_{i}}^{2}$. Thus, the total number of non-zero triangles with these points as vertices is $\sum_{i=1}^{12} \mathrm{C}_{x_{i}}^{2}$.
Therefore, the number of zero triangles is $\mathrm{C}_{12}^{3}-\sum_{i=1}^{12} \mathrm{C}_{x_{i}}^{2}$.
And $\sum_{i=1}^{12} x_{i}=\mathrm{C}_{12}^{2}=66$, so
$$
\sum_{i=1}^{12} \mathrm{C}_{x_{i}}^{2}=\frac{1}{2} \sum_{i=1}^{12}\left(x_{i}^{2}-x_{i}\right)=\frac{1}{2} \sum_{i=1}^{12} x_{i}^{2}-33 \text {. }
$$

By the adjustment method, it is known that when $\left\{x_{1}, x_{2}, \cdots, x_{12}\right\}$ has six taking 5 and six taking 6, $\sum_{i=1}^{12} \mathrm{C}_{x_{i}}^{2}$ has the minimum value of 150.

Therefore, the maximum number of zero triangles with these 12 points as vertices is 70.