5. In the Cartesian coordinate system, $F_{1}, F_{2}$ are the left and right foci of the hyperbola $\Omega: x^{2}-\frac{y^{2}}{3}=1$, respectively. A line $l$ passing through $F_{1}$ intersects $\Omega$ at two points $P, Q$. If $\overrightarrow{F_{1} F_{2}} \cdot \overrightarrow{F_{1} P}=16$, then the value of $\overrightarrow{F_{2} P} \cdot \overrightarrow{F_{2} Q}$ is $\qquad$.
Let $P(\sec \theta, \sqrt{3} \tan \theta), F_{1}(-2,0)$, then $\overrightarrow{F_{1} F_{2}}=(4,0)$,
$$
\begin{array}{l}
\overrightarrow{F_{1} P}=(\sec \theta+2, \sqrt{3} \tan \theta) \\
\Rightarrow \overrightarrow{F_{1} F_{2}} \cdot \overrightarrow{F_{1} P}=4(\sec \theta+2)=16 \Rightarrow \sec \theta=2
\end{array}
$$
$\Rightarrow \cos \theta=\frac{1}{2} \Rightarrow P(2, \pm 3)$. Without loss of generality, take $P(2,3)$,
Solving the system $\left\{\begin{array}{l}y=\frac{3}{4}(x+2), \\ 3 x^{2}-y^{2}=3\end{array} \Rightarrow 13 x^{2}-12 x-28=0\right.$
$\Rightarrow Q\left(-\frac{14}{13}, \frac{9}{13}\right)$, so $\overrightarrow{F_{2} P} \cdot \overrightarrow{F_{2} Q}=\left|\overrightarrow{F_{2} P}\right| \cdot\left|\overrightarrow{F_{2} Q_{1}}\right|=3 \cdot \frac{9}{13}=\frac{27}{13}$.