Convert the point $(\rho,\theta,\phi) = \left( 12, \frac{7 \pi}{6}, \frac{\pi}{3} \right)$ in spherical coordinates to rectangular coordinates.
For $x$, I get $x = 12 \sin \frac{\pi}{3} \cos \frac{7 \pi}{6} = 12 \cdot \frac{\sqrt{3}}{2} \cdot \frac{-\sqrt{3}}{2} = -9$. For $y$, I get $y = 12 \sin \frac{\pi}{3} \sin \frac{7 \pi}{6} = 12 \cdot \frac{\sqrt{3}}{2} \cdot \frac{-1}{2} = -3 \sqrt{3}$. For $z$, I get $z = 12 \cos \frac{\pi}{3} = 12 \cdot \frac{1}{2} = 6$. Therefore, the rectangular coordinates of the point are $(-9, -3 \sqrt{3}, 6)$. 