If   is the normal form of the equation of a straight line $x+\sqrt{3}y+4=0$and a, b are respectively x, y-intercepts of this line, then $\sqrt{3}\pi bp-3a\alpha =$

Given line

$\begin{array}{l}x+\sqrt{3}y+4=0\\ \Rightarrow x+\sqrt{3}y=-4\Rightarrow \frac{x}{-4}+\frac{y}{\left(\raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$\sqrt{3}$}\right.\right)}=1\\ a=-4,b=-\frac{4}{\sqrt{3}}and\\ x+\sqrt{3}y+4=0\Rightarrow x+\sqrt{3}y=-4\\ Dividewith\sqrt{a^2+b^2}=\sqrt{1+3}=2\\ \therefore \frac{x}{2}+\frac{\sqrt{3}}{2}y=\frac{-4}{2}\\ \Rightarrow \left(\frac{-1}{2}\right)x+\left(\frac{\sqrt{3}}{2}\right)y=2\\ \cos \alpha =\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\sin \alpha =\raisebox{1ex}{$-\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,andp=2\\ \alpha \in Q_3\Rightarrow \alpha =\pi +\frac{\pi }{3}=\frac{4\pi }{3}\\ Now\sqrt{3}\pi bp-3a\alpha =\sqrt{3}\pi \left(\frac{-4}{\sqrt{3}}\right)2-3\left(-4\right)\left(\frac{4\pi }{3}\right)\\ =-8\pi +16\pi \Rightarrow 8\pi \end{array}$