$A$ circle $C$ of radius $1$ is inscribed in an equilateral triangle $PQR$. The points of contact of $C$ with the sides $PQ, QR, RP$ are $D, E, F$,respectively. The line $PQ$ is given by the equation $\sqrt{3}x + y - 6 = 0$ and the point $D$ is $\left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)$. Further,it is given that the origin and the centre of $C$ are on the same side of the line $PQ$.
$1.$ The equation of circle $C$ is
$(A) (x - 2\sqrt{3})^2 + (y - 1)^2 = 1$
$(B) (x - 2\sqrt{3})^2 + (y + \frac{1}{2})^2 = 1$
$(C) (x - \sqrt{3})^2 + (y + 1)^2 = 1$
$(D) (x - \sqrt{3})^2 + (y - 1)^2 = 1$
$2.$ Points $E$ and $F$ are given by
$(A) \left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right), (\sqrt{3}, 0)$
$(B) \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right), (\sqrt{3}, 0)$
$(C) \left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right), \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$(D) \left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right), \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$3.$ Equation of the sides $QR, RP$ are
$(A) y = \frac{2}{\sqrt{3}}x + 1, y = -\frac{2}{\sqrt{3}}x - 1$
$(B) y = \frac{1}{\sqrt{3}}x, y = 0$
$(C) y = \frac{\sqrt{3}}{2}x + 1, y = -\frac{\sqrt{3}}{2}x - 1$
$(D) y = \sqrt{3}x, y = 0$
Give the answer for questions $1, 2$ and $3$.

• A
  $D, C, B$
• B
  $D, A, D$
• C
  $D, A, D$
• D
  $B, C, A$