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Get the area of the triangle by Heron's Formula: $$\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{(12)(3)(4)(5)}=12\sqrt{5}$$ Use the area to find the height $AH$ with known base $BC$: $$Area=12\sqrt{5}=\frac{1}{2}bh=\frac{1}{2}(8)(AH)$$ $$AH=3\sqrt{5}$$ $$BH=\sqrt{A{B}^2-A{H}^2}=\sqrt{7^2-(3\sqrt{5}{)}^2}=2$$ $$CH=BC-BH=8-2=6$$ Apply the Angle Bisector Theorem on $△ACH$ and $△ABH$, we get $AP:PH=9:6$ and $AQ:QH=7:2$, respectively. To find $AP$, $PH$, $AQ$, and $QH$, apply variables, such that $AP:PH=9:6$ is $\frac{3\sqrt{5}-x}{x}=\frac{9}{6}$ and $AQ:QH=7:2$ is $\frac{3\sqrt{5}-y}{y}=\frac{7}{2}$. Solving them out, you will get $AP=\frac{9\sqrt{5}}{5}$, $PH=\frac{6\sqrt{5}}{5}$, $AQ=\frac{7\sqrt{5}}{3}$, and $QH=\frac{2\sqrt{5}}{3}$. Then, since $AP+PQ=AQ$ according to the Segment Addition Postulate, and thus manipulating, you get $PQ=AQ-AP=\frac{7\sqrt{5}}{3}-\frac{9\sqrt{5}}{5}$ = $$\boxed{\frac{8}{15}\sqrt{5}}$$