Three, (50 points)
Divide each side of $\triangle A B C$ into $n$ equal parts, and draw lines parallel to the sides through each division point. How many parallelograms are there in the resulting figure?
First, consider parallelograms whose sides are not parallel to $BC$. Extend the sides of such parallelograms to intersect $BC$ at four points on $BC$. In special cases, the second and third intersection points may coincide (i.e., one vertex of the parallelogram). If the points on $BC$ are sequentially denoted as
$$
B_{1}=B, B_{2}, B_{3}, \cdots, B_{n}, B_{n+1}=C,
$$

then each parallelogram whose sides are not parallel to $BC$ corresponds to an ordered quadruple (i.e., the indices of the four points)
$$
(i, j, k, h) \text {, where } 1 \leqslant i<j \leqslant k<h \leqslant n+1 .
$$

This is a one-to-one correspondence, so the number of such parallelograms is equal to the number of such quadruples.
Define the mapping
$$
(i, j, k, h) \rightarrow(i, j, k+1, h-1)
$$

This mapping is a one-to-one correspondence from the set
$$
X=\{(i, j, k, h)|1 \leqslant i<j \leqslant k<h \leqslant n+1\}
$$

to the set
$$
Y=\left\{\left(i, j, k^{\prime}, h^{\prime}\right) \mid 1 \leqslant i<j<k^{\prime}<h^{\prime} \leqslant n+2\right\}
$$

Thus, $|X|=|Y|$. The number $|Y|$ is clearly the number of four-element subsets of the set $\{1,2, \cdots, n+2\}$, which is $C_{n+2}^{4}$. Therefore, $|X|=C_{n+2}^{4}$.

Hence, there are $C_{n+2}^{4}$ parallelograms whose sides are not parallel to $BC$. Similarly, we can consider parallelograms whose sides are not parallel to $AB$ or $AC$. Therefore, the total number of parallelograms in the figure is $3 C_{n+2}^{4} + 2$.