Given the rectangular coordinates of point Q:
[tex]Q(-6\sqrt{3},-6)[/tex]
You need to remember that the form from rectangular oordinates to polar coordinates is:
[tex](x,y)\rightarrow(r,\theta)[/tex]
By definition:
[tex]\begin{gathered} r=\sqrt{x^2+y^2} \\ \\ \theta=tan^{-1}(\frac{y}{x}) \end{gathered}[/tex]
In this case, you can identify that:
[tex]\begin{gathered} x=-6\sqrt{3} \\ y=-6 \end{gathered}[/tex]
Then, you can determine that:
[tex]\begin{gathered} r=\sqrt{(-6\sqrt{3})^2+(-6)^2}=12 \\ \\ \theta=tan^{-1}(\frac{-6}{-6\sqrt{3}})=\frac{5\pi}{6} \end{gathered}[/tex]
Therefore, the polar coordinates are:
[tex](12,\frac{5\pi}{6})[/tex]
Hence, the answer is: Second option.
