5. Let $n$ be a given natural number not less than 5. If

in the plane there are $n$ lines, among which exactly 3 lines are pairwise parallel, and in addition, no two lines are parallel, and no three lines are concurrent. Then the number of intersection points of these lines is $\qquad$
5. $\frac{n^{2}-n-6}{2}$.

First, remove 2 lines $l_{1}$ and $l_{2}$ from the 3 parallel lines, leaving $n-2$ lines, which gives
$$
\mathrm{C}_{n-2}^{2}=\frac{(n-2)(n-3)}{2}
$$

intersection points. Adding line $l_{1}$, it intersects with $n-3$ lines, resulting in $n-3$ intersection points. Similarly, adding line $l_{2}$, we get another $n-3$ intersection points. Therefore, the total number of intersection points is
$$
\frac{(n-2)(n-3)}{2}+2(n-3)=\frac{n^{2}-n-6}{2} \text {. }
$$