Let the sum of the focal distances of the point P(4,3) on the hyperbola $H:\ \dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ be $8\sqrt{\dfrac{5}{3}}$. If for H, the length of the latus rectum is l and the product of the focal distances of the point P is m, then $9l^{2}+6m$ is equal to:

Given sum of focal distances: $SP_1 + SP_2 = 2ex = 8\sqrt{\frac{5}{3}}$
At $P(4,3)$ ⇒ $2e \cdot 4 = 8\sqrt{\frac{5}{3}}$
⇒ $e=\sqrt{\frac{5}{3}}$

Now, $b^2 = a^2(e^2 - 1) = a^2\left(\frac{5}{3}-1\right)=\frac{2}{3}a^2$

Using point $(4,3)$: $\frac{16}{a^2}-\frac{9}{b^2}=1$
Substitute $b^2=\frac{2}{3}a^2$:
$\frac{16}{a^2}-\frac{9}{(2/3)a^2}=1$
⇒ $\frac{16}{a^2}-\frac{27}{2a^2}=1$
⇒ $\frac{5}{2a^2}=1$
⇒ $a^2=\frac{5}{2},\ b^2=\frac{5}{3}$

Latus rectum: $l=\frac{2b^2}{a}$ ⇒ $l^2=\frac{4b^4}{a^2}$
⇒ $l^2=\frac{4(25/9)}{5/2}=\frac{40}{9}$
⇒ $9l^2=40$

Product of focal distances: $m=(ex+a)(ex-a)=e^2x^2-a^2$
⇒ $m=\frac{5}{3}\cdot16-\frac{5}{2}=\frac{145}{6}$
⇒ $6m=145$

$9l^2+6m=40+145=185$

$\boxed{185}$