A *projective plane* is a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) $H=(V,E)$ such that


1. if $e\_1\neq e\_2 \in E$ then $|e\_1\cap e\_2| = 1$, and
2. for $v,w\in V$ there is $e\in E$ such that $\{v,w\}\subseteq e$.


Is there a projective plane $H=(V,E)$ such that



> 
> $|e|>2$ for all $e\in E$, and there are $e\_1, e\_2 \in E$ with $|e\_1|\neq |e\_2|$?
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> 
> 



Non-pencil infinite projective plane with edges of different cardinalities
Suppose $e\_1, e\_2$ are two distinct hyperedges, and $v$ is a vertex that is not a member of either of them. Define the map $f\_v: e\_1\to e\_2$ by sending a vertex $w\in e\_1$ to the unique vertex $w'\in e\_2$ which lies in the same hyperedge as $\{v,w\}$. The defining properties of projective planes imply that $f\_v$ is a bijection, therefore $|e\_1|=|e\_2|$.


