= Solution
Use the upper-half-plane model of the <hyperbolic plane>, whose <geodesics> are vertical lines and semicircles perpendicular to the real boundary. Hyperbolic and Euclidean angles agree. An isometry takes one of the <ultraparallel hyperbolic lines> to the imaginary axis. The other then has two finite endpoints on the same side; reflecting if necessary, write them as $0<a<b$. Its Euclidean centre and radius are $c=(a+b)/2$ and $r=(b-a)/2$.
A geodesic perpendicular to the imaginary axis must be a semicircle centred at zero, of some radius $q$. Two intersecting Euclidean circles meet orthogonally exactly when their squared centre distance equals the sum of their squared radii. Orthogonality to the second line therefore requires
$$
c^2=q^2+r^2,\qquad \boxed{q=\sqrt{ab}.}
$$
Since $a<\sqrt{ab}<b$, the two semicircles intersect at one point in the upper half-plane. This constructs the <common perpendicular of ultraparallel hyperbolic lines> and proves uniqueness, because its radius is forced.
For the three-line continuation, \b[the three common perpendiculars can be pairwise disjoint, but need not be]. Explicit <configurations of three ultraparallel lines> make both claims transparent. For the positive example, take the three semicircles of radius $1/2$ centred at $-5/2,0,5/2$. Their endpoint intervals are disjoint. Their common perpendiculars are the semicircles with centres $-5/4,5/4,0$ and radii $\sqrt{21}/4,\sqrt{21}/4,\sqrt6$, respectively. The two smaller semicircles are disjoint because $2(\sqrt{21}/4)<5/2$; each is nested inside the largest because $5/4+\sqrt{21}/4<\sqrt6$. Hence their full geodesics are disjoint. For the negative example, take three concentric semicircles of radii $1,2,3$. They are pairwise ultraparallel, but every common perpendicular is the imaginary axis, so the perpendiculars coincide.
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