The hyperbola is $\frac{x^2}{9}-\frac{y^2}{16}=1$, with vertex $(3,0)$. The line $8x-3y=24$ passes through $(3,0)$ and intersects the hyperbola again at $\left(5,\frac{16}{3}\right)$ (substitute $y=\frac{8x-24}{3}$).

The enclosed region lies between the right branch of the hyperbola and the line, from $x=3$ to $x=5$.

$A={\int }_3^5\left(\frac{4}{3}\sqrt{x^2-9}-\frac{8x-24}{3}\right)dx$. The triangular subtraction term integrates to $\frac{1}{2}\cdot 2\cdot \frac{16}{3}=\frac{16}{3}$.

Using $\int \sqrt{x^2-9}dx=\frac{x}{2}\sqrt{x^2-9}-\frac{9}{2}\ln \left|x+\sqrt{x^2-9}\right|+C$, evaluation from $3$ to $5$ gives $\frac{5}{2}(4)-\frac{9}{2}\ln 9+\frac{9}{2}\ln 3=10-\frac{9}{2}\ln 3$.

Hence $A=\frac{4}{3}\left(10-\frac{9}{2}\ln 3\right)-\frac{16}{3}=\frac{40-16}{3}-6\ln 3=8-6\ln 3$.

Therefore $A+6\ln 3=8$ and $3(A+6\ln 3)=24$.