Let $L =\{y=0\}\cap\mathcal{H}^{2}$, and let $P=(3,2,2)$. Show that the parallel postulate fails in $\mathcal{H}^{2}$ by giving two lines $L',L'' \in \mathcal{H}^{2}$ with $P\in L',L''$ and $L\cap L'=L\cap L'' = \varnothing$.
I think a good idea would be to drop a perpendicular from $P$ onto $L$. So I have found this line will lie on $\Pi \cap \mathcal{H}^{2}$, where $\Pi$ is the plane given by $(t−1)−2x−3y=0$. I am really not sure how to proceed from here.