smc

$ABC$ has side length $3$. Let $AP=A^{\prime }P=x$ and $AQ=A^{\prime }Q=y$. Thus, $BP=3-x$ and $CQ=3-y$. Applying Law of Cosines on triangles $BP{A}^{\prime }$ and $CQ{A}^{\prime }$ using the $60^{\circ }$ angles gives $x=\frac{7}{5}$ and $y=\frac{7}{4}$. Applying Law of Cosines once again on triangle $APQ$ using the $60^{\circ }$ angle gives $$P{Q}^2=\frac{(21)(49)}{400}$$ so $$PQ=\frac{7}{20}\sqrt{21}$$ The correct answer is $\boxed{\frac{7}{20}\sqrt{21}}$.