Given line,
$$
\begin{array}{l}
\sqrt{3} x+y=1 \\
y=-\sqrt{3} x+c
\end{array}
$$
We know that, $m=-\sqrt{3}$
$$
\begin{aligned}
\text { that, } m &=-\sqrt{3} \quad[\because y=m x+c] \\
\tan \theta &=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\
\tan 60^{\circ} &=\left|\frac{m_{1}-(-\sqrt{3})}{1+m_{1}(-\sqrt{3})}\right| \quad\left[\because \theta=60^{\circ}\right] \\
\sqrt{3}=&\left|\frac{m_{1}+\sqrt{3}}{1-\sqrt{3}} m_{1}\right| \\
\pm \sqrt{3}=& \frac{m_{1}+\sqrt{3}}{1-\sqrt{3} m_{1}}
\end{aligned}
$$
\begin{array}{l|l}
\hline taking + sign & taking - sign \\
\hline+\sqrt{3}=\frac{m_{1}+\sqrt{3}}{1-\sqrt{3} m_{1}} & -\sqrt{3}=\frac{m_{1}+\sqrt{3}}{1-\sqrt{3} m_{1}} \\
\sqrt{3}\left(1-\sqrt{3} m_{1}\right)=m_{1}+\sqrt{3} & \Rightarrow-\sqrt{3}\left(1-\sqrt{3} m_{1}\right) \\
& =m_{1}+\sqrt{3} \\
\Rightarrow \sqrt{3}-3 m_{1}=m_{1}+\sqrt{3} & \Rightarrow-\sqrt{3}-3 m_{1}=m_{1}+\sqrt{3} \\
\Rightarrow & \Rightarrow 3 m_{1}-m_{1}=2 \sqrt{3} \\
\Rightarrow 4m_{1}=0 & \Rightarrow \quad 2 m_{1}=2 \sqrt{3} \\
\Rightarrow m_{1}=0 & \Rightarrow \quad m_{1}=\sqrt{3}
\end{array}
given point (3,-2)
$y=m x$ + c
$$
\begin{aligned}
-2 &=3 \sqrt{3}+c \\
c &=-2-3 \sqrt{3}
\end{aligned}
$$
Hence, required equation
$$
\therefore \quad y=m x+c
$$
$$
\begin{array}{l}
y=\sqrt{3} x-2-3 \sqrt{3} \\
y-\sqrt{3} x+2+3 \sqrt{3}=0 \\
y-x \sqrt{3}+2+3 \sqrt{3}=0
\end{array}
$$