The equation of the straight line passing through the point $(3,2)$ and inclined at an angle of ${60}^{\circ }$with the line $\sqrt{3}x+y=1$ is 

Let m be the slope of required line, then

 $\tan {60}^{\circ }=\left|\frac{m-(-\sqrt{3})}{1+m(-\sqrt{3})}\right|$

$\begin{array}{l}\Rightarrow \sqrt{3}(1-m\sqrt{3})=\pm (m+\sqrt{3})\end{array}$

$\Rightarrow m=0$ or $m=\sqrt{3}$

As the line intersects the x-axis, $m\ne 0$ Thus, $m=\sqrt{3}$.

Equation of required line is

$y+2=\sqrt{3}(x-3)$ or $y-\sqrt{3}x+2+3\sqrt{3}=0$.