A straight line throughQ( $\sqrt{3}$,2) makes an angle $\frac{\mathrm{\pi }}{6}$ with the positive direction of the x-axis. If the straight line intersects the line $\sqrt{3}\mathrm{x}-4\mathrm{y}+8=0$ at P, find the distance PQ.

$\text{ Let }\mathrm{\theta }\text{ be the inclination of the line. Then }\mathrm{\theta }=\frac{\mathrm{\pi }}{6}\text{. }$
$\text{ Equation to the line passing through }\mathrm{Q}(\sqrt{3},2)$
$\text{ in Symmetric form is }\frac{\mathrm{x}-\sqrt{3}}{\cos \frac{\mathrm{\pi }}{6}}=\frac{\mathrm{y}-2}{\sin \frac{\mathrm{\pi }}{6}}=\mathrm{r}$
$\therefore \text{ any point on the line }=\left(\sqrt{3}+\mathrm{rcos}\frac{\mathrm{\pi }}{6},2+\mathrm{rsin}\frac{\mathrm{\pi }}{6}\right)$
$$\begin{gathered} \text{ Let this point be }\mathrm{P} \\ \therefore \mathrm{P}=\left(\sqrt{3}+\mathrm{r}\frac{\sqrt{3}}{2},2+\frac{\mathrm{r}}{2}\right) \end{gathered}$$
$\text{ Since }\mathrm{P}\text{ lies on the line }\mathrm{x}\sqrt{3}-4\mathrm{y}+8=0$
$\begin{array}{l}\Rightarrow \frac{3}{2}(2+\mathrm{r})-2(4+\mathrm{r})+8=0\\ \Rightarrow 3(2+\mathrm{r})-4(4+\mathrm{r})+16=0\Rightarrow 6-\mathrm{r}=0\\ \Rightarrow \mathrm{r}=6\Rightarrow \mathrm{PQ}=6\end{array}$