Find the equation of the locus of points equidistant from the lines $3x + 4y - 11 = 0$ and $12x + 5y + 2 = 0$.

The locus of the points which are at an equal distance from $3x + 4y - 11 = 0$ and $12x + 5y + 2 = 0$ and which is near the origin is:

The lines $L_1: y-x=0$ and $L_2: 2x+y=0$ intersect the line $L_3: y+2=0$ at $P$ and $Q$ respectively. The bisector of the angle between $L_1$ and $L_2$ divides the line segment $PQ$ internally at $R$.
Statement-$I$: $PR:RQ = 2\sqrt{2}:\sqrt{5}$
Statement-$II$: In any triangle,the bisector of an angle divides the opposite side in the ratio of the sides containing the angle.

Statement $-I$: Two lines which pass through a given fixed point and are equally inclined to two other lines passing through the same point,are always perpendicular to each other.
Statement $-II$: Angle bisectors of two intersecting lines are always perpendicular to each other.

The equation of the bisector of the angle between the lines $x + 2y - 11 = 0$ and $3x - 6y - 5 = 0$ which contains the point $(1, -3)$ is

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