1. **State the problem:** Convert the polar coordinates $\left(6\sqrt{2}, \frac{3\pi}{2}\right)$ into rectangular coordinates $(x,y)$. 2. **Recall the formulas:** For polar coordinates $(r, \theta)$, the rectangular coordinates are given by: $$x = r \cos \theta$$ $$y = r \sin \theta$$ 3. **Substitute the given values:** $$x = 6\sqrt{2} \cos \left(\frac{3\pi}{2}\right)$$ $$y = 6\sqrt{2} \sin \left(\frac{3\pi}{2}\right)$$ 4. **Evaluate the trigonometric functions:** $$\cos \left(\frac{3\pi}{2}\right) = 0$$ $$\sin \left(\frac{3\pi}{2}\right) = -1$$ 5. **Calculate $x$ and $y$:** $$x = 6\sqrt{2} \times 0 = 0$$ $$y = 6\sqrt{2} \times (-1) = -6\sqrt{2}$$ 6. **Final answer:** The rectangular coordinates are: $$\boxed{(0, -6\sqrt{2})}$$