Let $O(0, 0)$, $P(3, 4)$, $Q(6, 0)$ be the vertices of the triangles $OPQ$. The point $R$ inside the triangle $OPQ$ is such that the triangles $OPR$, $PQR$, $OQR$ are of equal area. The coordinates of $R$ are

$\text{Given: } O(0,0),\quad P(3,4),\quad Q(6,0)$ $\text{These are the vertices of } \triangle OPQ.$ $\text{Given: } \text{area}(\triangle OPR) = \text{area}(\triangle PQR) = \text{area}(\triangle OQR)$ $\text{To find: coordinates of } R$ $1.\ \text{find total area} = \frac{1}{2}\times OQ \times h$ $2.\ [\triangle OQR] = 4 \Rightarrow y$ $3.\ [\triangle OPR] = 4 \Rightarrow x$ $[\triangle OPQ] = \frac{1}{2}\times OQ \times h = \frac{1}{2}\times 6 \times 4$ $= 12$ $[\triangle OPR] = [\triangle PQR] = [\triangle OQR]$ $= \frac{12}{3} = 4$ $[\triangle OQR] = \frac{1}{2}\times OQ \times y = \frac{1}{2}\times 6 \times y$ $= 3y$ $3y = 4 \Rightarrow y = \frac{4}{3}$ $OP = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$ $\text{slope of } OP = \frac{4}{3}$ $\Rightarrow OP:\ 4x - 3y = 0$ $d = \frac{|4x - 3y|}{\sqrt{4^2+3^2}} = \frac{|4x-3y|}{5}$ $[\triangle OPR] = \frac{1}{2}\cdot OP\cdot d = \frac{1}{2}\cdot 5\cdot \frac{|4x-3y|}{5}$ $= \frac{|4x-3y|}{2} = 4 \Rightarrow |4x-3y| = 8$ $R \text{ inside } \triangle OPQ \Rightarrow 4x - 3y > 0$ $4x - 3y = 8$ $4x - 3\left(\frac{4}{3}\right) = 8 \Rightarrow 4x - 4 = 8$ $4x = 12 \Rightarrow x = 3$ $R = \left(3,\ \frac{4}{3}\right)$ $R = \text{centroid} = \left(\frac{0+3+6}{3},\ \frac{0+4+0}{3}\right)$ $= \left(3,\ \frac{4}{3}\right)$ $R = \left(3,\ \frac{4}{3}\right) \Rightarrow \text{option (3)}$