If the line $y=\sqrt{3}x-3$  cuts the parabola $y^2=x+2$  at P and Q if A be the point $(\sqrt{3},0)$  then $|AP.AQ|$  is

The point $A(\sqrt{3},0)$ lies on the given line whose slope is $\sqrt{3}$ . 

$\therefore \tan \theta =\sqrt{3}=\tan {60}^0\Rightarrow \theta ={60}^0$equation  of the line through A is
$\frac{x-\sqrt{3}}{\cos {60}^0}=\frac{y-0}{\sin {60}^0}=r$
Where r is the distance of any point from A in the direction of  ${60}^0$. It interests the parabola
$\begin{array}{l}{y}^2=x+2\Rightarrow {\left(\frac{r\sqrt{3}}{2}\right)}^2=\frac{r}{2}+\sqrt{3}+2\\ \Rightarrow 3r^2-2r-4(2+\sqrt{3})=0\\ r_1{r}_2=-\frac{4(2+\sqrt{3})}{3}=AP.AQ\\ \therefore AP.AQ=-\frac{4}{3}(2+\sqrt{3})\end{array}$