By the Law of Cosines, $\angle B A C=\cos ^{-1} \frac{3+1-7}{2 \sqrt{3}}=\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)=150^{\circ}$. If we let $Q$ be the intersection of $\ell_{2}$ and $A C$, we notice that $\angle Q B A=90^{\circ}-\angle Q A B=90^{\circ}-30^{\circ}=60^{\circ}$. It follows that triangle $A B P$ is a 30-60-90 triangle and thus $P B=2$ and $P A=\sqrt{3}$. Finally, we have $\angle P A C=360^{\circ}-\left(90^{\circ}+150^{\circ}\right)=120^{\circ}$, and $$P C=\left(P A^{2}+A C^{2}-2 P A \cdot A C \cos 120^{\circ}\right)^{1 / 2}=(3+3+3)^{1 / 2}=3$$
Triangle $A B C$ has $A B=1, B C=\sqrt{7}$, and $C A=\sqrt{3}$. Let $\ell_{1}$ be the line through $A$ perpendicular to $A B, \ell_{2}$ the line through $B$ perpendicular to $A C$, and $P$ the point of intersection of $\ell_{1}$ and $\ell_{2}$. Find $P C$.