Together with @TarasBanakh we faced the problem described in the title. Let me start with definitions.

A linear space is a pair $(S,\mathcal L)$ consisting of a set $S$ and a family $\mathcal L$ of subsets of $S$ where for any distinct points $x,y\in S$ there exists a unique line $L\in\mathcal L$ containing $x$ and $y$. We'll denote this line as $\overline{xy}$.

A hull of a set $A$ we will call a smallest set $\overline{A}$ such that $A \subset \overline{A} \land \forall x,y \in \overline{A} (x \not= y \rightarrow \overline{xy} \subset \overline{A}$). Having this hull definition we can easily define dimensional-based structures. In order to simplify writing, we will just write $\overline{a_1a_2...a_n}$ instead of $\overline{\{a_1, a_2, ..., a_n\}}$.

We'll call a set $A$ collinear if $\forall x, y \in A(x\not= y \rightarrow \overline{A} = \overline{xy})$. Set $P$ is called a plane if exist 3 non-collinear points $x, y, z\in S:P=\overline{xyz}$. Similarly we can define set planar if hull of any non-collinear 3-subset of it is the same plane $P$. At last, we'll call a set $S$ 3-dimensional space if there exist 4 distinct non-collinear and non-planar points $x,y,z,w \in S:S=\overline{xyzw}$.

We'll call 3-dimensional space balanced (due to BIBD) when cardinality of all lines is equal and cardinality of all planes is equal as well.

And at last, we'll call a 3-dimensional space $S$ Lobachevsky if for every plane $P\subset S$, line $L \subset P$ and point $x \in P \setminus L$ there exist at least two distinct lines $\Lambda_1, \Lambda_2$ in $S$ such that $x\in \Lambda_i\subset P\setminus L$. At last we can formulate the problem.

Problem. Does there exist finite balanced 3-dimensional Lobachevsky space $S$?

Some comments. Obviously there exist infinite any-dimensional Lobachevsky space. Finite affine and projective spaces are also well-known. Finite balanced Lobachevsky planes also exist (simplest example). If we need Lobachevsky plane where there exist no Pasch configuration, then we have unitals. So, we can see that there are infinitely-many balanced Lobachevsky planes. Still, I was not able to find any example of Lobachevsky 3-dimensional space.

Main difficulty is that such spaces should have lots of points. General formula for point number is very probably the following: $|S|=1 + (|L|-1)(1 + (k + |L| - 1) + (k + |L| - 1)^2)$ where $|L|\geq 3$ and $k\geq 2$ are number of points in line and number of "parallel" lines respectively (still could be wrong). Also not all $k$ are suitable. For instance, for 3-point lines only $k \equiv 0, 1 (\mod 3)$ are possible. According to this calculations, in case of 3-point lines, smallest example should have $63 = 1 + 2 * (1 + 5 + 25)$ points, next should have $87=1 + 2 * (1 + 6 + 36)$. If we're talking about 4-point lines, smallest should have $172 = 1 + 3 * (1 + 7 + 49)$, next $220 = 1 + 3 * (1 + 8 + 64)$ points. Therefore finding it by generative methods is very complicated (for instance, I checked all non-isomorphic $BIBD(63,3,1)$ generated by difference families and lots of similar $BIBD(87,3,1)$). Despite Lobachevsky 13-point and 15-point planes are difference-family-based, there is no difference-family-based $BIBD(63,3,1)$ and very probably no difference-family-based $BIBD(87,3,1)$.

I also tried to generate 3-dimensional analogues to unitals, Denniston maximal arcs, derived Steiner Triple Systems from Steiner Quadruple Systems and so on. I spent like 3 months of searching by computer and in literature/articles without any success. This is the motivation for problem above.