Non-pencil infinite projective plane with edges of different cardinalities

Question

A projective plane is a 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 $|e_1|\neq |e_2|$?

Answer 1

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 $|e_1|=|e_2|$.