Let $$(a, b)$$ be the point of intersection of the curve $$x^2 = 2y$$ and the straight line $$y - 2x - 6 = 0$$ in the second quadrant. Then the integral $$I = \int_{a}^{b} \frac{9x^2}{1 + 5^x}\,dx$$ is equal to :

The point of intersection of the parabola $$x^{2}=2y$$ and the straight line $$y-2x-6=0$$ is obtained by substituting $$y=2x+6$$ into the parabola.

$$x^{2}=2(2x+6)=4x+12$$
$$x^{2}-4x-12=0$$

Solving the quadratic gives $$x=\frac{4 \pm \sqrt{16+48}}{2}=6,-2$$.
In the second quadrant $$x\lt 0$$ and $$y\gt 0$$, so we choose $$x=-2$$.
Then $$y=2(-2)+6=2$$. Hence $$(a,b)=(-2,2)$$ and the integral limits are $$a=-2$$, $$b=2$$.

Define $$I=\int_{-2}^{2}\frac{9x^{2}}{1+5^{x}}\,dx$$ $$-(1)$$.

To exploit symmetry, substitute $$x\rightarrow -x$$ in $$I$$.
Because $$dx\rightarrow -dx$$, the limits interchange and the integral becomes

$$I=\int_{-2}^{2}\frac{9x^{2}}{1+5^{-x}}\,dx$$ $$-(2)$$.

Add $$(1)$$ and $$(2)$$:

$$2I=\int_{-2}^{2}9x^{2}\left[\frac{1}{1+5^{x}}+\frac{1}{1+5^{-x}}\right]dx$$.

Simplify the bracket. Let $$t=5^{x}$$, so $$5^{-x}=1/t$$:

$$\frac{1}{1+5^{x}}+\frac{1}{1+5^{-x}}=\frac{1}{1+t}+\frac{1}{1+1/t}=\frac{1}{1+t}+\frac{t}{1+t}=1$$.

Thus $$2I=\int_{-2}^{2}9x^{2}\,dx=9\int_{-2}^{2}x^{2}\,dx$$.

Compute the remaining integral:

$$\int_{-2}^{2}x^{2}\,dx=\left.\frac{x^{3}}{3}\right|_{-2}^{2}=\frac{8}{3}-\left(-\frac{8}{3}\right)=\frac{16}{3}$$.

Therefore $$2I=9\left(\frac{16}{3}\right)=48$$ and $$I=24$$.

The required value is $$24$$, which matches Option A.