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A finite linear space consists of a finite set $P$ of points and a set family $\mathcal{L}$ of subsets called lines, each containing at least two points, such that each pair of distinct points lies on exactly one line. Here the nontrivial convention also excludes a line equal to all of $P$. Two distinct lines therefore meet in at most one point, and counting pairs givesNear-pencils and finite projective planes, including the Fano plane, are examples. The De Bruijn--Erdos pair-covering inequality concerns the corresponding point degrees. The original line-count theorem proves $|\mathcal{L}|\ge |P|$, with equality only for a near-pencil or a finite projective plane.^[ref]

An exactly one-intersecting family with empty total intersection gives such a structure by using its members as points. Each original ground element supplies the set of members containing it. Discard the resulting sets of size zero or one; the others are lines, and every pair of member-points belongs to its unique shared-element line. Empty total intersection excludes an all-point line.

Yes. For every prime power $q\ge 2$, a finite projective plane of order $q$ haspoints and the same number of lines; every two lines meet in exactly one point, and the intersection of all lines is empty. Its lines therefore form one required family on $n$ points. The near-pencil from part (ii) is another. They are non-isomorphic because every projective-plane line has size $q+1$, whereas the near-pencil has a member of size $n-1$. There are infinitely many prime powers, so infinitely many such $n$.