The equation of two lines through the origin, which intersect  the line $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$ at angles of $\frac{\pi }{3}$ each, are

Given equation of line is

$\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$

$\Rightarrow$ DR's of the given line are $2, 1, 1$

 $\Rightarrow$ DC's of the given line are $\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}$

Since, required lines make an angle $\frac{\pi }{3}$ with the given line

The DC's of the required lines are

$\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{-1}{\sqrt{6}}\text{ and }\frac{-1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}}$ respectively. 

Also, both the required lines pass through the origin.

 :. Equation of required lines are 

$\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}\text{ and }\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}$