$3$ reds, $3$ blues and $3$ green marks to place to hexagon 

We have $3$ reds, $3$ blues and $3$ green marks to place on vertices in the given picture. (Only one mark for each vertex). The hexagon is free to move in three dimensions. How many nonequivalent configurations are there?

My try: If it were a line then there will be $9!/(3!)^3$ ways and by reflection we must also divide by $2$, but this is a hexagon. What should I do?
 A: By using Polya's theorem , we have two rotation $(\pi_0 ,\pi_1)$ and two reflections $(r_1,r_2)$ where $\pi_i=i \times 180$ , $i \in \{0,1\}$ and $r_1 = \text{reflection about vertical}$ , $r_2 = \text{reflection about horizontal}$ . Now , lets number the vertices on hexagon such that the leftmost vertex be $1$ and number consecutively as to clockwise such as $1,2,3,4,5,6$.Moreover, the remaining $3$ vertices on horizontal line be $7,8,9$ from left to right.

*

*$\pi_0 =\text{identity permutation}=(1)(2)...(8)(9)=x_1^9$


*$\pi_1 =\text{rotation about 180 degree}=(14)(25)(36)(79)(8)=x_1x_2^4$


*$r_1 =\text{horizontal reflection}=(26)(35)(1)(4)(7)(8)(9) =x_1^5x_2^2$


*$r_2 =\text{vertical reflection} = (14)(23)(65)(79)(8)=x_1x_2^4$
Let reds denoted by "x" , blues denoted by "y" , greens denoted by "z". Then , $$[x^3y^3z^3]\frac{1}{4}\bigg((x+y+z)^9 +2(x+y+z)(x^2+y^2+z^2)^4 +(x+y+z)^5 (x^2+y^2+z^2)^2\bigg)$$
EXPANSION OF FOREGOING EXPRESSION
So , the answer is $$\frac{1}{4} \times 1800 =450$$ where $1800$ is the coefficent of term $[x^3y^3z^3]$ in the expansion.
