Three. (This question is worth 50 points) Given 10 points in space, where no four points lie on the same plane. Connect some of the points with line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn.

Connect some of the points with line segments, if the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn.
Given 10 points as vertices and the lines connecting them as edges, we obtain a simple graph $G$ of order 10. We prove that the number of edges in $G$ does not exceed 15.

Let the vertices of $G$ be $v_{1}, v_{2}, \cdots, v_{10}$, with a total of $k$ edges. Let $\operatorname{deg}\left(v_{i}\right)$ denote the degree of vertex $v_{i}$. If $\operatorname{deg}\left(v_{i}\right) \leq 3$ for $i=1,2, \cdots, 10$, then
$$
k=\frac{1}{2} \sum_{i=1}^{10} \operatorname{deg}\left(v_{i}\right) \leq \frac{1}{2} \times 10 \times 3=15.
$$

Assume there exists a $v_{i}$ such that $\operatorname{deg}\left(v_{i}\right) \geq 4$. Without loss of generality, let $\operatorname{deg}\left(v_{1}\right)=n \geq 4$, and $v_{1}$ is adjacent to $v_{2}, \cdots, v_{n+1}$. Thus, there are no edges between $v_{2}, \cdots, v_{n+1}$, otherwise a triangle would be formed. Therefore, there are exactly $n$ edges between $v_{1}, v_{2}, \cdots, v_{n+1}$.

For each $j(n+2 \leq j \leq 10)$, $v_{j}$ can be adjacent to at most one of $v_{2}, v_{3}, \cdots, v_{n+1}$ (otherwise, if $v_{j}$ is adjacent to $v_{s}, v_{t}(2 \leq s<t \leq n+1)$, then $v_{1}, v_{s}, v_{j}, v_{t}$ would form a quadrilateral in space, contradicting the given condition), so the number of edges between $v_{2}, \cdots, v_{n+1}$ and $v_{n+2}, \cdots, v_{10}$ is at most $10-(n+1)=9-n$.
20 points
Among the $9-n$ vertices $v_{n+2}, \cdots, v_{n}$, since there are no triangles, by Turán's theorem, there are at most $\left[\frac{(9-n)^{2}}{4}\right]$ edges. Therefore, the number of edges in $G$ is
$$
k \leq n+(9-n)+\left[\frac{(9-n)^{2}}{4}\right]=9+\left[\frac{(9-n)^{2}}{4}\right] \leq 9+\left[\frac{25}{4}\right]=15. \cdots \cdots 30 \text{ points}
$$

The graph given in the figure has 15 edges and meets the requirements.
In conclusion, the maximum number of edges is 15.
.50 points