I am trying to figure out the least number of lines which guarantees that 3 will meet at the same point in a projective plane of order $n$. Through experimentation with $n=2$ and $3$ I get (unless I have overlooked something) $5$ and $6$ respectively. This is essentially by drawing pairs of parallel lines of varying 'gradients' till it seems a new line would cross a previous intersection. With $n+k$ lines we have $\frac{1}{2}(n+k)(n+k-1)$ intersections amongst $n^2+n+1$ possibilities, but I am having a hard time arguing what the pattern here is and why.
Is there a nice intuitive answer to this question?
edit for what it's worth, this question came about because in the game Spot It! (i.e. ${\sf {PG}(2,7)}$ with $2$ points removed) there is a variation which involves placing $9$ cards on the table and having players trying to find a three-way match (1 symbol on three cards). The value of $9$ is one lower than the 'easy bound', which got me wondering whether it is indeed a guarantee or not.
Pick the first line $\ell$ arbitrarily. How many lines can you add before some point on $\ell$ belongs to at least three lines?