jmc

Point $G$ is the centroid of $△DEF$, so $DG:GP=EG:GQ=2:1$. Therefore, $DG=\frac{2}{3}(DP)=12$ and $QG=\frac{1}{3}(EQ)=8$, so applying the Pythagorean Theorem to $△QGD$ gives us $QD=\sqrt{Q{G}^2+G{D}^2}=\sqrt{64+144}=\sqrt{16(4+9)}=4\sqrt{13}$, which means $DF=2QD=\boxed{8\sqrt{13}}$.