Step 1 :The rectangular coordinates of the point are given as $(-3 \sqrt{3}, 3)$.
Step 2 :We can find the polar coordinates $(r, \theta)$ of this point using the formulas $r = \sqrt{x^2 + y^2}$ and $\theta = \arctan(\frac{y}{x})$.
Step 3 :Substituting the given values into the formula for $r$, we get $r = \sqrt{(-3 \sqrt{3})^2 + 3^2} = 6$.
Step 4 :Substituting the given values into the formula for $\theta$, we get $\theta = \arctan(\frac{3}{-3 \sqrt{3}})$. However, since the point is in the second quadrant, we need to add 180 degrees to the result of the arctangent function, giving us $\theta = 150$ degrees.
Step 5 :Therefore, the polar coordinates of the point $(-3 \sqrt{3}, 3)$ are $(6, 150^\circ)$.
Step 6 :Final Answer: \(\boxed{(6, 150^\circ)}\)