Let $m$ be the slope of the line,

then $\tan {60}^o=\left|\frac{m+\sqrt{3}}{1-m\sqrt{3}}\right|$

$\Rightarrow \sqrt{3}=\left|\frac{m+\sqrt{3}}{1-\sqrt{3} m}\right|$

$\Rightarrow \pm \sqrt{3}=\frac{m+\sqrt{3}}{1-\sqrt{3} m}$

$\therefore \sqrt{3}-3m=m+\sqrt{3}$ or $-\sqrt{3}+3m=m+\sqrt{3}$

$4m=0$ or $2m=2\sqrt{3}$

$m=0$ or $m=\sqrt{3}$

Since L intersects $X$-axis we take $m=\sqrt{3}$

and get $y=\sqrt{3}x+c$

Since it passes through $\left(3,-2\right)$, therefore line will become $y-\sqrt{3}x+2+3\sqrt{3}=0$