Let the slope of the line 45x + 5y + 3 = 0 be 27 r _ 1 + 9 r _ 2 2...

We know that,

The slope of line $45x + 5y + 3 = 0$ will be $- 9$

Now, given $27{r}_{1} + \dfrac{9{r}_{2}}{2} = - 9$

Now, solving

$L = \lim_{x \to 3}\dfrac{{∫}_{3}^{x}8{t}^{2}dt}{\dfrac{3{r}_{2}x}{2} - {r}_{2}{x}^{2} - {r}_{1}{x}^{3} - 3x}$

Applying Newton Leibnitz Theorem and L'Hospital Rule we get,

$⇒ L = \lim_{x \to 3}\dfrac{8{x}^{2}}{\dfrac{3{r}_{2}^{}}{2} - 2{r}_{2}x - 3{r}_{1}{x}^{2} - 3}$

$⇒ L = \dfrac{72}{\dfrac{3{r}_{2}}{2} - 6{r}_{2} - 27{r}_{1} - 3}$

$⇒ L = \dfrac{72}{- \dfrac{9{r}_{2}}{2} - 27{r}_{1} - 3}$

Now, using $27{r}_{1} + \dfrac{9{r}_{2}}{2} = - 9$ we get,

$⇒ L = \dfrac{72}{9 - 3} = 12$