Equation of Straight Line Inclined at Sixty Degrees

To find the equation of the straight line $L$, we first determine the slope of the given line:
$\sqrt{3}x + y = 1 \implies y = -\sqrt{3}x + 1$

The slope of this line is $m_1 = -\sqrt{3}$.
Thus, its angle of inclination $\theta_1$ with the positive direction of the $x$-axis is:
$\tan \theta_1 = -\sqrt{3} \implies \theta_1 = 120^\circ$

The line $L$ is inclined at an angle of $60^\circ$ to the given line. Therefore, the angle of inclination $\theta$ of line $L$ can be:
$\theta = \theta_1 \pm 60^\circ$

This gives two possible inclinations for $L$:

1. $\theta = 120^\circ + 60^\circ = 180^\circ \equiv 0^\circ$
2. $\theta = 120^\circ - 60^\circ = 60^\circ$

Case 1: If $\theta = 0^\circ$:
The slope of line $L$ is $m = \tan 0^\circ = 0$.
The equation of the line passing through $(3, -2)$ with slope $0$ is:
$y - (-2) = 0(x - 3) \implies y = -2$
This line is parallel to the $x$-axis and hence does not intersect the $x$-axis. Since the problem explicitly states that $L$ intersects the $x$-axis, this case is rejected.

Case 2: If $\theta = 60^\circ$:
The slope of line $L$ is $m = \tan 60^\circ = \sqrt{3}$.
The equation of the line passing through $(3, -2)$ with slope $\sqrt{3}$ is:
$y - (-2) = \sqrt{3}(x - 3)$
$y + 2 = \sqrt{3}x - 3\sqrt{3}$
$y - \sqrt{3}x + 2 + 3\sqrt{3} = 0$

This line intersects the $x$-axis, satisfying all conditions.

Therefore, the correct option is (B).