The image of the centroid of the triangle whose ortho centre is at origin and the circum centre at $\left(3,3\right)$  in the line which is equally inclined with axes in the first quadrant is 

In any triangle the centroid $G$ divides the line segment joining ortho center  $O$and the circum centre $S$ in the ratio  $2:1$
It implies 
 $\begin{array}{c}G=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)\\ =\left(\frac{2\left(3\right)+1\left(0\right)}{1+2},\frac{2\left(3\right)+1\left(0\right)}{1+2}\right)\\ =\left(2,2\right)\end{array}$ 
Equation of the line which is equally inclined, means making equal angles with axes in the first quadrant is  $x=y$
The image of the point $P\left(x_1,y_1\right)$ in the line   $x=y$is   $\left(y_1,x_1\right)$
The image of  $G=\left(2,2\right)$ in the line  $x=y$ is  $\boxed{\left(2,2\right)}$