1. Interpret the given line correctly

The printed line is intended as $\sqrt{3}\,x + y = 1,$ not $\sqrt{3x}+y=1$.

So its slope is $m_1=-\sqrt{3}.$

2. Use angle formula for slopes

Let the slope of the required line $L$ be $m$. Since $L$ makes an angle $60^\circ$ with the line $\sqrt{3}x+y=1$, we use $\tan\theta=\left|\frac{m-m_1}{1+mm_1}\right|.$ Here $\theta=60^\circ$ and $m_1=-\sqrt{3}$, so $\left|\frac{m+\sqrt{3}}{1-m\sqrt{3}}\right|=\sqrt{3}.$

This gives two cases:

Case 1

$\frac{m+\sqrt{3}}{1-m\sqrt{3}}=\sqrt{3}$ $m+\sqrt{3}=\sqrt{3}-3m$ $4m=0$ $m=0.$

This is the x-axis direction, i.e. the line through $(3,-2)$ would be $y=-2,$ which does not intersect the x-axis. So reject this.

Case 2

$\frac{m+\sqrt{3}}{1-m\sqrt{3}}=-\sqrt{3}$ $m+\sqrt{3}=-\sqrt{3}+3m$ $2\sqrt{3}=2m$ $m=\sqrt{3}.$

So the required line has slope $m=\sqrt{3}.$

3. Equation through $(3,-2)$

Using point-slope form, $y+2=\sqrt{3}(x-3).$

Expanding, $y=\sqrt{3}x-3\sqrt{3}-2.$

Bringing all terms to one side, $y-\sqrt{3}x+2+3\sqrt{3}=0.$

4. Match with the options

This is exactly Option B: $y-\sqrt{3}x+2+3\sqrt{3}=0.$

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.