7. Given five points in a plane, where no two lines connecting any two points are parallel, perpendicular, or coincident, draw perpendiculars from one of the points to the lines connecting the other points. How many intersection points can these perpendiculars have, excluding the original five points, at most?

保留源文本的换行和格式，所以翻译结果如下：

7. Given five points in a plane, where no two lines connecting any two points are parallel, perpendicular, or coincident, draw perpendiculars from one of the points to the lines connecting the other points. How many intersection points can these perpendiculars have, excluding the original five points, at most?
7. For five points, the number of lines connecting any two points is $\mathrm{C}_{5}^{2}=10$, and the number of triangles is $\mathrm{C}_{5}^{3}=10$.

Among any four points, the number of lines connecting them is $\mathrm{C}_{4}^{2}=6$, and the fifth point can draw 6 perpendicular lines to these lines, resulting in a total of 30 perpendicular lines. If we do not consider non-intersecting and repeated lines, we can get $\mathrm{C}_{30}^{2}=435$ intersection points.

For the five points, there are 10 connecting lines, and each line has three parallel perpendicular lines, so we need to subtract 30 intersection points.
In the 10 triangles, the three altitudes of each triangle intersect at one point, so we should subtract $\left(\mathrm{C}_{3}^{2}-1\right) \times 10=$ 20 intersection points.

Furthermore, for each point $P_{i}$, the perpendicular lines drawn to $\mathrm{C}_{4}^{2}=6$ lines intersect at the same point $P_{i}$ (excluding $P_{i}$ itself), so we should subtract $\mathrm{C}_{6}^{2} \times 5=75$, leaving us with $435-30-20-75=310$ intersection points.