Think between the lines

Suppose $n-1$ lines have already been drawn.

Drawn the $n$ th line so that it intersects each of the existing lines once, avoiding the existing intersections. In doing so, you divide $n$ existing regions in two, thus creating $n$ new regions.

If an existing region was closed, it is divided into two closed regions; if an existing region was open, it is now divided into an open and a closed region, unless it is the very first or the very last region, in which case both parts remain open. In short, you will create $n-2$ new closed regions.

It is easy to see that $N_0 = N_1 = N_2 = 0$ , and $N_n = N_{n-1} + (n-2)$ ; thus $N_n$ is the $(n-2)$ th triangle number, $N(n) = \tfrac12(n-2)(n-1) = \tfrac12(n^2 - 3n + 2).$

This would be much better poised as an open-ended question (if only Brilliant.org had such a feature....)

Given the options given, and knowinf that you can come up with only one triangle given 3 lines, you substitute $n=3$ into the equations and see which of the formulas give a value of 1. Luckily, it turns out that only one of the options gives this result.