The area of the region included between the parabola$y=\frac{3x^2}{4}$ and the line $3x-2y+12=0$ is

 

$y=\frac{3x^2}{4}\to 1$

$3x-2y+12=0\to 2$

$1\&2\Rightarrow 3x-2\left(\frac{3x^2}{4}\right)+12=0$

$-\frac{3x^2}{4}+\frac{3x}{2}+\frac{12}{2}=0$

$R.A=\frac{{\Delta }^{3/2}}{6a^2}=\frac{{\left(\frac{81}{4}\right)}^{3/2}}{6\left(\frac{9}{16}\right)}$

                      $=\frac{{\left(\frac{9}{2}\right)}^3}{6.\left(\frac{9}{16}\right)}$

                     $=27$