Equations of Sides of Equilateral Triangle

To find the equations of the sides $QR$ and $RP$ of the equilateral triangle $PQR$, we proceed through the following steps:

Step 1: Find the Centre of the Inscribed Circle $C$

The line equation of side $PQ$ is:
$\sqrt{3}x + y - 6 = 0$

The unit normal vector to the line $PQ$ directed outwards is:
$\hat{\mathbf{n}} = \left(\frac{\sqrt{3}}{\sqrt{(\sqrt{3})^2 + 1^2}}, \frac{1}{\sqrt{(\sqrt{3})^2 + 1^2}}\right) = \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$

The point of contact on side $PQ$ is given as $D = \left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)$ and the radius of the circle is $r = 1$. The centre of the circle $O'$ lies along the normal to the line $PQ$ passing through $D$:
$O' = D \pm r \hat{\mathbf{n}} = \left(\frac{3\sqrt{3}}{2} \pm \frac{\sqrt{3}}{2}, \frac{3}{2} \pm \frac{1}{2}\right)$

We are given that the origin $(0, 0)$ and the centre $O'$ lie on the same side of the line $PQ$.
Substituting $(0, 0)$ into the line equation:
$\sqrt{3}(0) + 0 - 6 = -6 < 0$

Choosing the minus sign for $O'$:
$O' = \left(\frac{3\sqrt{3}}{2} - \frac{\sqrt{3}}{2}, \frac{3}{2} - \frac{1}{2}\right) = (\sqrt{3}, 1)$
Evaluating the line equation at $(\sqrt{3}, 1)$:
$\sqrt{3}(\sqrt{3}) + 1 - 6 = 3 + 1 - 6 = -2 < 0$
Since this has the same sign as the origin, the centre of the circle $C$ is indeed:
$O' = (\sqrt{3}, 1)$

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Step 2: Determine the Slopes of the Sides $QR$ and $RP$

The slope of side $PQ$ is:
$m_1 = -\sqrt{3} \implies \theta_1 = 120^\circ$

Since $\triangle PQR$ is an equilateral triangle, each interior angle is $60^\circ$. Hence, the angles of inclination of the other two sides $QR$ and $RP$ must differ from $\theta_1$ by $\pm 60^\circ$:
$\theta_2 = 120^\circ - 60^\circ = 60^\circ \implies m_2 = \tan 60^\circ = \sqrt{3}$
$\theta_3 = 120^\circ + 60^\circ = 180^\circ \equiv 0^\circ \implies m_3 = \tan 0^\circ = 0$

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Step 3: Find the Equations of the Tangents

The circle has centre $(\sqrt{3}, 1)$ and radius $r = 1$. Both sides $QR$ and $RP$ are tangent to this circle.

1. For the side with slope $m_3 = 0$:
   The equation is horizontal:
   $y = c$
   The perpendicular distance from the centre $(\sqrt{3}, 1)$ to this line must be equal to the radius $r = 1$:
   $|1 - c| = 1 \implies 1 - c = \pm 1 \implies c = 0 \text{ or } c = 2$
   To form a valid triangle containing the inscribed circle with side $PQ$ (where $y \le 6$), the tangent is:
   $y = 0$

2. For the side with slope $m_2 = \sqrt{3}$:
   The equation is of the form:
   $\sqrt{3}x - y + c' = 0$
   The perpendicular distance from $(\sqrt{3}, 1)$ to this line is:
   $\frac{|\sqrt{3}(\sqrt{3}) - 1 + c'|}{\sqrt{(\sqrt{3})^2 + (-1)^2}} = 1 \implies \frac{|3 - 1 + c'|}{2} = 1 \implies |2 + c'| = 2$
   This gives:
   $2 + c' = \pm 2 \implies c' = 0 \text{ or } c' = -4$
   For $c' = 0$, the line passes through the origin and intersects the other two sides to form the equilateral triangle surrounding the circle:
   $\sqrt{3}x - y = 0 \implies y = \sqrt{3}x$

Thus, the equations of the sides $QR$ and $RP$ are:
$y = \sqrt{3}x \quad \text{and} \quad y = 0$

Therefore, the correct option is D.