A ray of light coming from the point $$(2, 2\sqrt{3})$$ is incident at an angle $$30^\circ$$ on the line $$x = 1$$ at the point A. The ray gets reflected on the line $$x = 1$$ and meets $$x$$-axis at the point B. Then, the line AB passes through the point:

We have a source point $$S(2,\,2\sqrt3)$$. The vertical mirror is the straight line $$x = 1$$. Let the point of incidence on the mirror be $$A(1,\,y)$$, where $$y$$ is to be determined.

The incident ray is the segment joining $$S$$ and $$A$$, so its slope is obtained in the usual way:

$$m_{\text{inc}}=\frac{y-2\sqrt3}{1-2}=2\sqrt3-y.$$

The question states that the ray strikes the line $$x=1$$ at an angle of $$30^\circ$$. Because the line $$x=1$$ is vertical, “angle with the mirror” means the angle between the ray and the vertical line. Hence the same ray makes an angle of $$90^\circ-30^\circ=60^\circ$$ with the positive $$x$$-axis (i.e. with the horizontal). For a line which makes an angle $$\theta$$ with the positive $$x$$-axis, the magnitude of its slope is

$$|m|=\tan\theta.$$

Putting $$\theta=60^\circ$$ we obtain

$$|m_{\text{inc}}|=\tan60^\circ=\sqrt3.$$ Therefore

$$|\,2\sqrt3-y\,|=\sqrt3.$$

This single modulus equation splits into two ordinary equations:

1. $$2\sqrt3-y=\sqrt3\qquad\Longrightarrow\qquad y=\sqrt3,$$

2. $$2\sqrt3-y=-\sqrt3\qquad\Longrightarrow\qquad y=3\sqrt3.$$

Thus the point of incidence can be either $$A_1(1,\sqrt3)\quad\text{or}\quad A_2(1,3\sqrt3).$$

Next we impose the law of reflection: the angle of incidence equals the angle of reflection. For a vertical mirror this simply reverses the sign of the horizontal component of any direction vector, so the slope of the reflected ray is the negative of the incident slope, i.e.

$$m_{\text{ref}}=-m_{\text{inc}}.$$

We now treat the two candidate points separately to see which one produces a reflected ray that passes through one of the points given in the options.

Case 1: $$A_1(1,\sqrt3).$$

For this point $$m_{\text{inc}}=2\sqrt3-\sqrt3=\sqrt3,$$ so $$m_{\text{ref}}=-\sqrt3.$$ The equation of the reflected ray through $$A_1$$ is written in point-slope form:

$$y-\sqrt3=-\sqrt3\,(x-1).$$

We test the four candidate points one by one.

For $$\bigl(3,-\sqrt3\bigr):$$ $$\text{LHS}=y-\sqrt3=-\sqrt3-\sqrt3=-2\sqrt3,$$ $$\text{RHS}=-\sqrt3\,(3-1)=-2\sqrt3,$$ so the point satisfies the equation. Hence line $$A_1B$$ (with $$m=-\sqrt3$$) indeed passes through $$\bigl(3,-\sqrt3\bigr).$$

Because a single straight line cannot pass through two non-collinear points, the other three options need not be checked further. Therefore the reflected ray that starts at $$A_1$$ includes the point $$\bigl(3,-\sqrt3\bigr).$$

Case 2: $$A_2(1,3\sqrt3).$$

For this point $$m_{\text{inc}}=2\sqrt3-3\sqrt3=-\sqrt3,$$ so $$m_{\text{ref}}=\sqrt3.$$ The equation of the reflected ray is

$$y-3\sqrt3=\sqrt3\,(x-1).$$

Substituting each of the four option points shows that none of them satisfies this second equation, so Case 2 yields no admissible answer.

Consequently, only the ray that reflects from $$A_1(1,\sqrt3)$$ is consistent with the data and the given options, and that ray passes through the point $$\bigl(3,-\sqrt3\bigr).$$

Hence, the correct answer is Option C.