A straight line $$L$$ through the point $$(3, -2)$$ is inclined at an angle of $$60^\circ$$ to the line $$\sqrt{3}x + y = 1$$. If $$L$$ also intersects the X-axis, then the equation of $$L$$ is:

We begin by examining the line already given in the question. The equation $$\sqrt{3}x + y = 1$$ can be rewritten in the slope-intercept form $$y = mx + c$$ in order to read its slope directly. Subtracting $$\sqrt{3}x$$ from both sides we get

$$y = -\sqrt{3}\,x + 1.$$

So, the slope of this reference line is $$m_1 = -\sqrt{3}.$$

The new line $$L$$ must make an angle of $$60^\circ$$ with this reference line. For two lines whose slopes are $$m_1$$ and $$m_2$$, the magnitude of the angle $$\theta$$ between them is given by the standard formula

$$\tan\theta \;=\;\left|\dfrac{m_2 - m_1}{1 + m_1\,m_2}\right|.$$

Here $$\theta = 60^\circ$$ and thus $$\tan 60^\circ = \sqrt{3}.$$ Putting these values in, we have

$$\Bigl|\dfrac{m_2 - (-\sqrt{3})}{1 + (-\sqrt{3})\,m_2}\Bigr| \;=\; \sqrt{3}.$$

Simplifying the numerator inside the absolute value, we obtain

$$\Bigl|\dfrac{m_2 + \sqrt{3}}{\,1 - \sqrt{3}\,m_2\,}\Bigr| \;=\; \sqrt{3}.$$

This absolute-value equation splits into two linear equations:

1. $$\dfrac{m_2 + \sqrt{3}}{1 - \sqrt{3}\,m_2} = \sqrt{3},$$ 2. $$\dfrac{m_2 + \sqrt{3}}{1 - \sqrt{3}\,m_2} = -\sqrt{3}.$$

First possibility. Take $$\dfrac{m_2 + \sqrt{3}}{1 - \sqrt{3}\,m_2} = \sqrt{3}.$$ Cross-multiplying:

$$(m_2 + \sqrt{3}) = \sqrt{3}\,(1 - \sqrt{3}\,m_2).$$

Expanding the right side gives

$$m_2 + \sqrt{3} = \sqrt{3} - 3\,m_2.$$

Bringing all terms to the left:

$$m_2 + 3\,m_2 + \sqrt{3} - \sqrt{3} = 0 \;\Longrightarrow\; 4\,m_2 = 0 \;\Longrightarrow\; m_2 = 0.$$

A slope of zero corresponds to a horizontal line $$y = \text{constant}.$$ Passing this horizontal line through the given point $$(3,-2)$$ would give $$y = -2,$$ which never meets the X-axis (where $$y = 0$$). Therefore this solution is inadmissible.

Second possibility. Now take $$\dfrac{m_2 + \sqrt{3}}{1 - \sqrt{3}\,m_2} = -\sqrt{3}.$$ Cross-multiplying again:

$$(m_2 + \sqrt{3}) = -\sqrt{3}\,(1 - \sqrt{3}\,m_2).$$

Expanding the right side gives

$$m_2 + \sqrt{3} = -\sqrt{3} + 3\,m_2.$$

Collecting like terms on the left:

$$$ \begin{aligned} m_2 - 3\,m_2 + \sqrt{3} + \sqrt{3} &= 0 \\ -2\,m_2 + 2\sqrt{3} &= 0. \end{aligned} $$$

Dividing by $$-2$$ we get

$$m_2 = \sqrt{3}.$$

Hence the slope of the required line $$L$$ is $$m_2 = \sqrt{3}.$$

Because $$L$$ passes through the point $$(3,-2)$$, we use the point-slope form $$y - y_1 = m(x - x_1).$$ Substituting $$x_1 = 3,\; y_1 = -2$$ and $$m = \sqrt{3}$$, we write

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

Removing the double negative:

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

Now we bring every term to the left to obtain the standard (general) form:

$$$ \begin{aligned} y + 2 - \sqrt{3}\,x + 3\sqrt{3} &= 0 \\ y - \sqrt{3}\,x + 2 + 3\sqrt{3} &= 0. \end{aligned} $$$

This is already one of the listed options. We may still verify that the line indeed intersects the X-axis. Setting $$y = 0$$ in the equation gives

$$0 - \sqrt{3}\,x + 2 + 3\sqrt{3} = 0 \;\Longrightarrow\; \sqrt{3}\,x = 2 + 3\sqrt{3} \;\Longrightarrow\; x = \dfrac{2 + 3\sqrt{3}}{\sqrt{3}},$$

which is a real number, so the line meets the X-axis exactly once. All conditions are satisfied.

The expression $$y - \sqrt{3}x + 2 + 3\sqrt{3} = 0$$ appears as Option D.

Hence, the correct answer is Option D.