$$\begin{aligned} &\sqrt{(c+4)^2+9}+\sqrt{(c-4)^2+9}=8 \sqrt{\frac{5}{3}}\\ &\text { Solving, } c=\frac{5}{\sqrt{6}}=\mathrm{al} \Rightarrow a^2\left(1+\frac{b^2}{a^2}\right)=\frac{25}{6} \Rightarrow a^2+b^2=\frac{25}{6}\\ &\begin{aligned} & \frac{16}{a^2}-\frac{9}{b^2}=1 \\ & 16 b^2-9 d^2=a^2 b^2 \Rightarrow 16\left(\frac{25}{6}-a^2\right)-9 a^2=9 a^2 b^2 \\ & P F_1+P F_2=8 \sqrt{\frac{5}{3}} \Rightarrow a^2=\frac{5}{2}, b^2=\frac{5}{3} \\ & \left|P F_1-P F_2\right|=2 a \\ & \frac{64.5}{3}=4 a^2+4 m \Rightarrow m=\frac{80}{3}-a^2 \\ & 6 m=160-6 a^2 \end{aligned} \end{aligned}$$

$$\begin{aligned} & 9 \ell^2=9\left(\frac{2 b^2}{a}\right)^2=\frac{36 b^4}{a^2} \\ & 9 \ell^2+6 m=\frac{36\left(\frac{25}{9}\right)}{\frac{5}{2}}+160-6\left(\frac{5}{2}\right) \\ & =\frac{72 \times 5}{9}+160-15 \\ & =160+40-15=185 \end{aligned}$$