If the area of the region bounded by $$16x^2 - 9y^2 = 144$$ and $$8x - 3y = 24$$ is $$A$$, then $$3(A + 6\log_e(3))$$ is equal to _________.

The hyperbola $$16x^2-9y^2=144$$ can be written as $$\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$$, whose right branch lies in $$x\ge 3$$ because the vertex is at $$(\pm 3,0)$$.

The straight line is $$8x-3y=24 \;\Longrightarrow\; y=\dfrac{8x-24}{3}=\dfrac83\,(x-3).$$

To obtain the bounded region, find the points where the two curves meet:
Substitute $$y=\dfrac{8x-24}{3}$$ in $$16x^2-9y^2=144$$:

$$\begin{aligned} 16x^2-9\left(\dfrac{8x-24}{3}\right)^2 &= 144\\ 16x^2-(8x-24)^2 &= 144\\ 16x^2-\left(64x^2-384x+576\right) &= 144\\ -48x^2+384x-576 &= 144\\ 48x^2-384x+720 &= 0\\ x^2-8x+15 &= 0\\ (x-3)(x-5) &= 0 \end{aligned}$$

Hence intersection points are $$(3,0)$$ and $$\left(5,\dfrac{16}{3}\right).$$ For $$3\lt x\lt 5$$ the hyperbola’s upper branch $$y=\dfrac{4}{3}\sqrt{x^2-9}$$ lies above the line, giving a closed region bounded by these two curves between $$x=3$$ and $$x=5$$.

Area $$A$$ of this region:

$$\begin{aligned} A &= \int_{3}^{5}\left[\dfrac43\sqrt{x^2-9}-\dfrac{8x-24}{3}\right]dx\\[3pt] &= \dfrac13\int_{3}^{5}\Bigl[4\sqrt{x^2-9}-8x+24\Bigr]dx. \end{aligned}$$

Split the integral:

$$I_1 = 4\int\sqrt{x^2-9}\,dx, \qquad I_2 = \int(-8x+24)\,dx.$$

Standard result: $$\displaystyle\int\sqrt{x^2-a^2}\,dx=\dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\ln\!\left|x+\sqrt{x^2-a^2}\right|.$$ Here $$a=3$$, so

$$\begin{aligned} I_1 &=4\left[\dfrac{x}{2}\sqrt{x^2-9}-\dfrac{9}{2}\ln\!\left(x+\sqrt{x^2-9}\right)\right]\\ &= 2x\sqrt{x^2-9}-18\ln\!\left(x+\sqrt{x^2-9}\right). \end{aligned}$$

For $$I_2$$: $$\displaystyle\int(-8x+24)\,dx=-4x^2+24x.$$

Thus

$$\begin{aligned} A &= \dfrac13\Bigl[\,2x\sqrt{x^2-9}-18\ln\!\left(x+\sqrt{x^2-9}\right)-4x^2+24x\Bigr]_{3}^{5}. \end{aligned}$$

Evaluate at the upper limit $$x=5$$ (note $$\sqrt{25-9}=4$$):

$$2(5)(4)-18\ln(5+4)-4(25)+24(5)=40-18\ln 9-100+120=60-18\ln 9.$$

Evaluate at the lower limit $$x=3$$ (note $$\sqrt{9-9}=0$$):

$$0-18\ln(3+0)-36+72=36-18\ln 3.$$

Difference:

$$60-18\ln 9-(36-18\ln 3)=24-18(\ln 9-\ln 3).$$

But $$\ln 9=2\ln 3$$, hence $$\ln 9-\ln 3=\ln 3$$ and

$$24-18\ln 3.$$

Finally,

$$A=\dfrac13\left(24-18\ln 3\right)=8-6\ln 3.$$

Required expression:

$$\begin{aligned} 3\bigl(A+6\ln 3\bigr) &= 3\bigl[(8-6\ln 3)+6\ln 3\bigr]\\ &=3\times 8\\ &=24. \end{aligned}$$

Therefore the value is $$\boxed{24}$$.