I think the answer is yes simply because the only examples are finite projective planes.

Do you know otherwise? Is this what you suspect? If so, why not ask *that*?

Let me recast things in another way:
Say that there is a set of lines $\mathcal{L}=\{L_1,\cdots L_m\}$, a set of points $\mathcal{P}=\{p_1,\cdots,p_n\}$, and a relation of incidence between points and lines.

Your key requirement for a linear intersection structure (LIS) is:

- There are never two lines $L,L'$ and two points $p,p'$ with both points incident with both lines.

Because of the symmetry one sees that roles of points and lines can be swapped giving another LIS. We will not be bothered, though, by saying two lines intersect when we mean that there is a point incident to both.

Here are two stronger conditions, either of which implies that a system is a LIS:

Those which satisfy the first condition are called linear spaces (of the geometric variety) and are known to always have $n \leq m.$

Those satisfying the second could be called dual-linear spaces.

Here is a troublesome example satisfying both. A **pencil** consists of $n$ lines and $n$ points with $l_i$ incident with just $p_i$ and $p_n$ except that $l_n$ is incident with exactly $p_1,\cdots,p_{n-1}.$ Note that it is self dual. The coloring number is $n$ but the line size is not uniform.

One way to exclude this is, as you do, saying that every line is incident with at least $3$ points. A gentler one is to simply state that there are is some set of $4$ points no three co-linear. See if you can prove that that alone is enough to make all lines the same size (given linear and dual linear).

DO you have any bad examples in mind other than pencils which have line coloring number $n$ , non-uniform line size? If not, why not use the less exclusive definition.

For a linear space $n\leq m$ while for a dual linear space $m \leq n$ so for one satisfying both, $n=m.$

It is known that for spaces satisfying both conditions along with the $4$ point condition, there is a $q$ with $n=m=q^2+q+1,$ All lines are incident with $q+1$ points and each point is incident with $q+1$ lines.

SO if your conditions force a projective plane then, yes, all the lines have the same size.

To back up, The coloring number can be defined via a graph with $m$ nodes with edge $u_iu_j$ if the corresponding lines intersect. Your condition is that this graph has chromatic number at least $n.$ One way to force this is to have a clique of size $n$, i.e. $n$ pairwise intersecting lines. In that case we could drop any other lines and still have a LIS with coloring number $n$ although perhaps this is uniform while the larger one wasn't.

Here is something I feel like I have seen but I can't find any reference:

- A set of $n$ pairwise intersecting lines on $n$ points is a projective plane or a pencil.

This leaves the case of clique number less than $n$ and coloring number perhaps greater than $n$ but certainly no smaller.

Can this happen? I doubt it.

If we drop one incident point from one line in a projective plane the number of points is $n^2+n$ and the number of lines is still $n^2+n+1$ However the coloring number goes don to $n^2+n$

What can we say about graphs with clique number $\kappa$ and coloring number $\kappa+1$ or more? Specifically, what is a lower bound on the number of lines (we only care about ones coming from LIS but we can ignore that until we need it. )

In the case $\kappa=2$ there are graphs which are triangle-free but have arbitrarily high chromatic number. However the number of point grows quadratically with the coloring number.

Finally there are calculations one can do depending of the smallest and/or largest line size. I'll just say that when the smallest line size is $3$ or more than the number of lines can't excede $\frac{n(n-1)}6$