Answer:
Option B
Explanation:
A straight lire passing through P and making an angle of $\alpha=60^{0}$ is given by
$\frac{y-y_{1}}{x-x_{1}}= \tan ( \theta \pm \alpha)$
where $\sqrt{3}x+y=1$
$\Rightarrow$ $y=-\sqrt{3}x+1$
Then , $\tan \theta= -\sqrt{3}$
$\Rightarrow$ $\frac{y+2}{x-3}=\frac{\tan \theta \pm \tan \alpha}{ 1 \pm \tan \theta \tan \alpha}$
$\Rightarrow$ $\frac{y+2}{y-3}=\frac{-\sqrt{3}+\sqrt{3}}{1-(-\sqrt{3})(\sqrt{3})}$
and $\frac{y+2}{x-3}= \frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3})(\sqrt{3})}$
$\Rightarrow$ y+2=0
and $\frac{y+2}{x-3}=\frac{-2 \sqrt{3}}{1-3}= \sqrt{3}$
$\Rightarrow$ $y+2= \sqrt{3}x-3 \sqrt{3}$
Neglecting , y+2=0 as it does not interest Y-axis
