Step-by-step Derivation:

1. Find the Center of the Inscribed Circle

• Let the center of the circle C be O'(h, k). The radius is given as r = 1.
• The equation of the side PQ is $L: \sqrt{3}x + y - 6 = 0$. The slope of this line is $m_{PQ} = -\sqrt{3}$.
• The radius from the center O' to the point of contact D is perpendicular to the tangent line PQ.
• The coordinates of D are given as $(3\sqrt{3}/2, 3/2)$.
• The slope of the line segment O'D is $m_{O'D} = -1/m_{PQ} = -1/(-\sqrt{3}) = 1/\sqrt{3}$.
• The center O'(h, k) must lie on the line passing through D with slope $1/\sqrt{3}$. The equation of this line is: $y - \frac{3}{2} = \frac{1}{\sqrt{3}} \left(x - \frac{3\sqrt{3}}{2}\right)$ $y - \frac{3}{2} = \frac{x}{\sqrt{3}} - \frac{3}{2}$ $y = \frac{x}{\sqrt{3}} \implies x = \sqrt{3}y$ So, the coordinates of the center satisfy $h = \sqrt{3}k$.
• The distance between the center O'(h, k) and the point of contact D $(3\sqrt{3}/2, 3/2)$ is equal to the radius, r=1. $(h - \frac{3\sqrt{3}}{2})^2 + (k - \frac{3}{2})^2 = 1^2$
• Substitute $h = \sqrt{3}k$ into the distance formula: $(\sqrt{3}k - \frac{3\sqrt{3}}{2})^2 + (k - \frac{3}{2})^2 = 1$ $(\sqrt{3}(k - \frac{3}{2}))^2 + (k - \frac{3}{2})^2 = 1$ $3(k - \frac{3}{2})^2 + (k - \frac{3}{2})^2 = 1$ $4(k - \frac{3}{2})^2 = 1$ $(k - \frac{3}{2})^2 = \frac{1}{4} \implies k - \frac{3}{2} = \pm \frac{1}{2}$
• This gives two possible values for k:
  • k = 3/2 + 1/2 = 2. Then $h = \sqrt{3}(2) = 2\sqrt{3}$. Center is $(2\sqrt{3}, 2)$.
  • k = 3/2 - 1/2 = 1. Then $h = \sqrt{3}(1) = \sqrt{3}$. 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. We check this by evaluating the line equation $L(x, y) = \sqrt{3}x + y - 6$ at the origin and the two possible centers.
  • $L(0, 0) = \sqrt{3}(0) + 0 - 6 = -6$ (Negative).
  • For $(2\sqrt{3}, 2)$: $L(2\sqrt{3}, 2) = \sqrt{3}(2\sqrt{3}) + 2 - 6 = 6 + 2 - 6 = 2$ (Positive).
  • For $(\sqrt{3}, 1)$: $L(\sqrt{3}, 1) = \sqrt{3}(\sqrt{3}) + 1 - 6 = 3 + 1 - 6 = -2$ (Negative).
• Since $L(\sqrt{3}, 1)$ has the same sign as L(0, 0), the center of the circle C is $O' = (\sqrt{3}, 1)$.

2. Find the Slopes of the Other Sides

• The triangle PQR is equilateral, so the angle between any two sides is 60°.
• The slope of PQ is $m_1 = -\sqrt{3}$. Let the slope of another side be $m_2$.
• The angle $\theta$ between two lines is given by $\tan\theta = |(m_1 - m_2) / (1 + m_1 m_2)|$.
• $\tan(60°) = \sqrt{3} = |(-\sqrt{3} - m_2) / (1 - \sqrt{3}m_2)|$.
• This yields two equations:
  1. $\sqrt{3}(1 - \sqrt{3}m_2) = -\sqrt{3} - m_2 \implies \sqrt{3} - 3m_2 = -\sqrt{3} - m_2 \implies 2\sqrt{3} = 2m_2 \implies m_2 = \sqrt{3}$.
  2. $\sqrt{3}(1 - \sqrt{3}m_2) = -(-\sqrt{3} - m_2) \implies \sqrt{3} - 3m_2 = \sqrt{3} + m_2 \implies -4m_2 = 0 \implies m_2 = 0$.
• So, the slopes of the other two sides, QR and RP, are $\sqrt{3}$ and 0.

3. Find the Equations of Sides QR and RP

• For an equilateral triangle, the incenter O' is also the orthocenter. The line O'D is the altitude from vertex R to side PQ. Thus, vertex R lies on the line passing through O' and D, which we found to be $y = x/\sqrt{3}$.
• The distance from a vertex to the incenter in an equilateral triangle is the circumradius R. We know R = 2r. Since r = 1, R = 2.
• Let the coordinates of vertex R be $(x_R, y_R)$. The distance O'R = 2. $(x_R - \sqrt{3})^2 + (y_R - 1)^2 = 2^2 = 4$
• Since R lies on $x = \sqrt{3}y$, we substitute $x_R = \sqrt{3}y_R$: $(\sqrt{3}y_R - \sqrt{3})^2 + (y_R - 1)^2 = 4$ $3(y_R - 1)^2 + (y_R - 1)^2 = 4$ $4(y_R - 1)^2 = 4 \implies (y_R - 1)^2 = 1 \implies y_R - 1 = \pm 1$
• This gives two possible locations for R: $y_R = 2$ (so $x_R = 2\sqrt{3}$) or $y_R = 0$ (so $x_R = 0$).
• The incenter O' must lie between the vertex R and the side PQ (on which D lies). The y-coordinates on the altitude line are: $y_{O'} = 1$ and $y_D = 3/2$. The order of points on the altitude from R to PQ must be R, O', D. This implies $y_R < y_{O'} < y_D$ or $y_R > y_{O'} > y_D$. Since 1 < 3/2, we must have $y_R < 1$. The only solution is $y_R = 0$.
• Therefore, the vertex R is at the origin (0, 0).
• The sides QR and RP pass through R(0, 0) and have slopes $\sqrt{3}$ and 0.
  • Side with slope $\sqrt{3}$: $y - 0 = \sqrt{3}(x - 0) \implies y = \sqrt{3}x$.
  • Side with slope 0: $y - 0 = 0(x - 0) \implies y = 0$.
• The equations of the sides QR and RP are $y = \sqrt{3}x$ and y = 0.

4. Conclusion

The equations of the other two sides are $y = \sqrt{3}x$ and y = 0. This matches option D.