Step-by-step Solution:

1. Find the center of the inscribed circle C.

Let the center of the circle C be $O'=(h, k)$. The radius is given as $r=1$. The line PQ has the equation $L \equiv \sqrt{3}x + y - 6 = 0$. The point of contact is $D = \left( \frac{3\sqrt{3}}{2}, \frac{3}{2} \right)$.

The line segment $O'D$ is a radius of the circle and is perpendicular to the tangent line PQ. The slope of PQ is $m_{PQ} = -\sqrt{3}$. The slope of the normal line $O'D$ is $m_{O'D} = -1/m_{PQ} = 1/\sqrt{3}$.

The equation of the line passing through D and containing the center O' is: $y - \frac{3}{2} = \frac{1}{\sqrt{3}} \left(x - \frac{3\sqrt{3}}{2}\right)$ $y - \frac{3}{2} = \frac{1}{\sqrt{3}}x - \frac{3}{2}$ $y = \frac{x}{\sqrt{3}} \implies k = \frac{h}{\sqrt{3}}$ So, the center $(h,k)$ lies on the line $y=x/\sqrt{3}$.

The distance from the center $O'(h, k)$ to the line PQ is equal to the radius, $r=1$. $|\frac{\sqrt{3}h + k - 6}{\sqrt{(\sqrt{3})^2 + 1^2}}| = 1$ $|\frac{\sqrt{3}h + k - 6}{2}| = 1$ $\sqrt{3}h + k - 6 = 2 \quad \text{or} \quad \sqrt{3}h + k - 6 = -2$

Substitute $k = h/\sqrt{3}$ into these two equations:

Case 1: $\sqrt{3}h + \frac{h}{\sqrt{3}} = 8 \implies \frac{3h+h}{\sqrt{3}} = 8 \implies 4h = 8\sqrt{3} \implies h = 2\sqrt{3}$. Then $k = \frac{2\sqrt{3}}{\sqrt{3}} = 2$. The center is $(2\sqrt{3}, 2)$.

Case 2: $\sqrt{3}h + \frac{h}{\sqrt{3}} = 4 \implies \frac{4h}{\sqrt{3}} = 4 \implies h = \sqrt{3}$. Then $k = \frac{\sqrt{3}}{\sqrt{3}} = 1$. The center is $(\sqrt{3}, 1)$.

We are given that the origin $(0,0)$ and the center of C are on the same side of the line PQ. Let's check the sign of the expression $L(x,y) = \sqrt{3}x + y - 6$ at the origin and the two possible centers. At the origin: $L(0,0) = \sqrt{3}(0) + 0 - 6 = -6$, which is negative.

For Case 1, center $(2\sqrt{3}, 2)$: $L(2\sqrt{3}, 2) = \sqrt{3}(2\sqrt{3}) + 2 - 6 = 6+2-6 = 2$, which is positive. This is rejected. For Case 2, center $(\sqrt{3}, 1)$: $L(\sqrt{3}, 1) = \sqrt{3}(\sqrt{3}) + 1 - 6 = 3+1-6 = -2$, which is negative. This is accepted.

So, the center of the circle C is $O' = (\sqrt{3}, 1)$.

2. Find the coordinates of points E and F.

The points D, E, F are the points of contact of the incircle with the sides of an equilateral triangle. These points D, E, F themselves form an equilateral triangle inscribed in circle C. The angle between the radii to any two consecutive points of contact is $120^\circ$. We can find the coordinates of E and F by rotating the vector $\vec{O'D}$ by $\pm 120^\circ$ about the center O'.

The vector from the center O' to the point D is: $\vec{O'D} = D - O' = \left( \frac{3\sqrt{3}}{2} - \sqrt{3}, \frac{3}{2} - 1 \right) = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)$ Let this vector be represented in complex numbers as $z_D = \frac{\sqrt{3}}{2} + i\frac{1}{2} = e^{i \pi/6}$. To get the vectors $\vec{O'E}$ and $\vec{O'F}$, we rotate this by $\pm 120^\circ$ (or $\pm 2\pi/3$ radians).

Rotation by $+120^\circ$: $z_E = z_D \cdot e^{i 2\pi/3} = e^{i \pi/6} \cdot e^{i 2\pi/3} = e^{i (\pi/6 + 4\pi/6)} = e^{i 5\pi/6}$ $z_E = \cos(5\pi/6) + i \sin(5\pi/6) = -\frac{\sqrt{3}}{2} + i\frac{1}{2}$ This corresponds to the vector $\vec{O'E} = \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$. The coordinates of point E are $O' + \vec{O'E} = (\sqrt{3}, 1) + \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right) = \left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right)$.

Rotation by $-120^\circ$: $z_F = z_D \cdot e^{-i 2\pi/3} = e^{i \pi/6} \cdot e^{-i 2\pi/3} = e^{i (\pi/6 - 4\pi/6)} = e^{-i 3\pi/6} = e^{-i \pi/2}$ $z_F = \cos(-\pi/2) + i \sin(-\pi/2) = 0 - i$ This corresponds to the vector $\vec{O'F} = (0, -1)$. The coordinates of point F are $O' + \vec{O'F} = (\sqrt{3}, 1) + (0, -1) = (\sqrt{3}, 0)$.

So, the other two points of contact are E and F, with coordinates $\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right)$ and $(\sqrt{3}, 0)$.

3. Compare with options. The calculated points are $\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right)$ and $(\sqrt{3}, 0)$. This matches option A.

Final check: Point D: $\left( \frac{3\sqrt{3}}{2}, \frac{3}{2} \right)$ Point E: $\left( \frac{\sqrt{3}}{2}, \frac{3}{2} \right)$ Point F: $(\sqrt{3}, 0)$

Distance DE: $\sqrt{(\frac{3\sqrt{3}}{2} - \frac{\sqrt{3}}{2})^2 + (\frac{3}{2}-\frac{3}{2})^2} = \sqrt{(\sqrt{3})^2} = \sqrt{3}$. Distance EF: $\sqrt{(\frac{\sqrt{3}}{2} - \sqrt{3})^2 + (\frac{3}{2}-0)^2} = \sqrt{(-\frac{\sqrt{3}}{2})^2 + (\frac{3}{2})^2} = \sqrt{\frac{3}{4} + \frac{9}{4}} = \sqrt{\frac{12}{4}} = \sqrt{3}$. Distance FD: $\sqrt{(\sqrt{3} - \frac{3\sqrt{3}}{2})^2 + (0-\frac{3}{2})^2} = \sqrt{(-\frac{\sqrt{3}}{2})^2 + (-\frac{3}{2})^2} = \sqrt{\frac{3}{4} + \frac{9}{4}} = \sqrt{\frac{12}{4}} = \sqrt{3}$. Since all sides are equal, the points D, E, F form an equilateral triangle, as expected.