To convert from cartesian coordinates $(x,y,z)$ to cylindrical coordinates $(r,\theta,z)$, we use the following formulas:
$$r=\sqrt{x^2+y^2}, \quad \theta = \tan^{-1}(y/x), \quad z=z$$
Applying these formulas to the point $(9,-9,2)$, we get:
$$r=\sqrt{9^2+(-9)^2}=9\sqrt{2}$$
$$\theta=\tan^{-1}(-9/9)=-\frac{\pi}{4}+\pi n, \quad \text{where } n \in \mathbb{Z}$$
Note that we get a negative angle because the point is in the third quadrant. We can add any integer multiple of $\pi$ to get all possible values of $\theta$ for this point. Therefore, we can write:
$$\theta=-\frac{\pi}{4}+\pi n, \quad \text{where } n \in \mathbb{Z}$$
Finally, we have $z=2$. So the cylindrical coordinates of the point $(9,-9,2)$ are:
$$(r,\theta,z)=\boxed{(9\sqrt{2}, -\frac{\pi}{4}+\pi n, 2)}, \quad \text{where } n \in \mathbb{Z}$$
Therefore, the answer is none of the choices given.