Let $$P = (-1, 0), Q = (0, 0)$$ and $$R = (3, 3\sqrt{3})$$ be three points. The equation of the bisector of the angle PQR

For an angle bisector we need the two rays that form the angle and the vertex where they meet.
Here the vertex is $$Q(0,0)$$, the two rays are $$QP$$ and $$QR$$.

Step 1: Direction of $$QP$$
$$P(-1,0)$$ and $$Q(0,0)$$ give the vector $$\overrightarrow{QP} = (-1,0)$$.
This lies along the negative $$x$$-axis, so its direction makes an angle $$180^{\circ}$$ with the positive $$x$$-axis.
Slope of $$QP$$: $$m_1 = 0$$.

Step 2: Direction of $$QR$$
$$R(3,3\sqrt 3)$$ and $$Q(0,0)$$ give the vector $$\overrightarrow{QR} = (3,3\sqrt 3)$$.
Slope of $$QR$$: $$m_2 = \frac{3\sqrt 3}{3} = \sqrt 3$$.
The corresponding angle with the positive $$x$$-axis is $$\tan^{-1}(\sqrt 3)=60^{\circ}$$.

Step 3: Identify the interior angle at $$Q$$
Angles of the two rays measured from the positive $$x$$-axis are $$60^{\circ}$$ (for $$QR$$) and $$180^{\circ}$$ (for $$QP$$).
The smaller angle between them is $$180^{\circ}-60^{\circ}=120^{\circ}$$.
Hence the interior bisector will make half of this, i.e. $$\tfrac{120^{\circ}}{2}=60^{\circ}$$, with each ray.
Starting from $$60^{\circ}$$ and moving halfway towards $$180^{\circ}$$ takes us to
$$60^{\circ}+60^{\circ}=120^{\circ}$$.

Step 4: Slope of the interior bisector
Angle $$120^{\circ}$$ has slope $$\tan(120^{\circ}) = -\sqrt 3$$.
Therefore the bisector through the origin is
$$y = -\sqrt 3\,x$$.

Step 5: Write in general form
$$\sqrt 3\,x + y = 0$$.

Thus the equation of the bisector of $$\angle PQR$$ is
Option A which is: $$\sqrt{3}\,x + y = 0$$.