Singular cardinal $\kappa$ with projective plane such that all edges have cardinality $<\kappa$

Question

Is there an infinite singular cardinal $\kappa$ such that there is a set $E\subseteq{\cal P}(\kappa)$ with the following properties?

1. $|e| < \kappa$ for all $e\in E$,
2. whenever $\alpha\neq\beta\in \kappa$ there is $e\in E$ with $\{\alpha,\beta\} \subseteq e$, and
3. if $e_1\neq e_2\in E$ then $|e_1\cap e_2| = 1$.

(There can be no infinite regular cardinal with this property.)

Answer 1

The answer is no. Let $\kappa$ be any infinite cardinal, regular or singular, and assume for a contradiction that there is a set $E\subseteq\mathcal P(\kappa)$ satisfying your conditions. I will call the elements of $\kappa$ points and the elements of $E$ lines.

The lines do not all go through one point: Given a point $\alpha$, choose a point $\beta\ne\alpha$ and a point $\gamma$ not on the line through $\alpha$ and $\beta$; the line through $\beta$ and $\gamma$ does not go through $\alpha$.

There are $\lt\kappa$ lines through any point: Consider any point $\alpha$ and let $\lambda$ be the number of lines through $\alpha$. Choose a line $e$ which does not go through $\alpha$. Since each line through $\alpha$ meets $e$ in a different point, $\lambda\le|e|\lt\kappa$.

Now choose two distinct points $\alpha$ and $\beta$. Say there are $\lambda$ lines through $\alpha$ and $\mu$ lines through $\beta$. Let $e$ be the line through $\alpha$ and $\beta$. Now every point which is not on the line $e$ is the point of intersection of a line through $\alpha$ and a line through $\beta$. Hence $\kappa\le|e|+\lambda\mu\lt\kappa$ which is absurd.

This argument is adapted from the proof that a finite projective plane of order $n$ has $n^2+n+1$ points. In that case we have $|e|=\lambda=\mu=n+1$ and the number of points is exactly $|e|+(\lambda-1)(\mu-1)=n^2+n+1$.

P.S. The answer is still no if condition (2) is weakened to "for every $\alpha\in\kappa$ we have $|\{e\in E:\alpha\in e\}|\gt1$". This more general result was proved in my answer to this old question.