Let the slope of the line $$45x + 5y + 3 = 0$$ be $$27r_1 + \frac{9r_2}{2}$$ for some $$r_1, r_2 \in R$$. Then $$\lim_{x \to 3}\left(\int_3^x \frac{8t^2}{\frac{3r_2 x}{2} - r_2 x^2 - r_1 x^3 - 3x} dt\right)$$ is equal to ______.

We need to find $$r_1, r_2$$ from the slope of the line $$45x + 5y + 3 = 0$$, then evaluate the given limit.

$$45x + 5y + 3 = 0 \implies y = -9x - \frac{3}{5}$$

Slope = $$-9$$.

We are given: slope = $$27r_1 + \frac{9r_2}{2}$$

$$27r_1 + \frac{9r_2}{2} = -9$$

$$54r_1 + 9r_2 = -18$$

$$6r_1 + r_2 = -2 \quad \cdots (i)$$

This gives one equation with two unknowns. We need another condition from the limit expression.

$$\lim_{x \to 3}\left(\int_3^x \frac{8t^2}{\frac{3r_2 x}{2} - r_2 x^2 - r_1 x^3 - 3x}\, dt\right)$$

Note that the denominator of the integrand depends on $$x$$ but not on $$t$$. So we can factor it out:

$$= \lim_{x \to 3} \frac{1}{\frac{3r_2 x}{2} - r_2 x^2 - r_1 x^3 - 3x} \int_3^x 8t^2\, dt$$

$$\int_3^x 8t^2\, dt = \frac{8t^3}{3}\Big|_3^x = \frac{8}{3}(x^3 - 27)$$

Let $$g(x) = \frac{3r_2 x}{2} - r_2 x^2 - r_1 x^3 - 3x$$.

$$g(3) = \frac{9r_2}{2} - 9r_2 - 27r_1 - 9 = -\frac{9r_2}{2} - 27r_1 - 9$$

$$= -(27r_1 + \frac{9r_2}{2} + 9) = -(-9 + 9) = 0$$

using equation (i) where $$27r_1 + \frac{9r_2}{2} = -9$$.

So we have a $$\frac{0}{0}$$ form. We apply L'Hopital's rule (differentiating with respect to $$x$$).

Numerator: $$\frac{d}{dx}\left[\frac{8}{3}(x^3 - 27)\right] = \frac{8}{3} \cdot 3x^2 = 8x^2$$

Denominator: $$g'(x) = \frac{3r_2}{2} - 2r_2 x - 3r_1 x^2 - 3$$

$$g'(3) = \frac{3r_2}{2} - 6r_2 - 27r_1 - 3 = -\frac{9r_2}{2} - 27r_1 - 3$$

$$= -(27r_1 + \frac{9r_2}{2}) - 3 = -(-9) - 3 = 9 - 3 = 6$$

$$\lim_{x \to 3} \frac{8x^2}{g'(x)} = \frac{8(9)}{6} = \frac{72}{6} = 12$$

The correct answer is 12.