1. **State the problem:** Find the distance between the two parallel planes given by the equations: $$z=2y-2x$$ and $$3z=-7-6x+6y$$ 2. **Rewrite both planes in standard form:** For the first plane: $$z=2y-2x \implies z - 2y + 2x = 0$$ For the second plane: $$3z = -7 - 6x + 6y \implies 3z + 6x - 6y = -7$$ Divide the entire second equation by 3 to simplify: $$z + 2x - 2y = -\frac{7}{3}$$ 3. **Check if planes are parallel:** The normal vector of the first plane is: $$\vec{n_1} = (2, -2, 1)$$ The normal vector of the second plane is: $$\vec{n_2} = (2, -2, 1)$$ Since $\vec{n_1} = \vec{n_2}$, the planes are parallel. 4. **Formula for distance between parallel planes:** $$\text{Distance} = \frac{|D_2 - D_1|}{\|\vec{n}\|}$$ where the planes are in the form: $$Ax + By + Cz + D = 0$$ and $\vec{n} = (A, B, C)$. 5. **Identify constants $D_1$ and $D_2$:** First plane: $$2x - 2y + z + 0 = 0 \implies D_1 = 0$$ Second plane: $$2x - 2y + z + \frac{7}{3} = 0 \implies D_2 = \frac{7}{3}$$ 6. **Calculate the magnitude of the normal vector:** $$\|\vec{n}\| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$$ 7. **Calculate the distance:** $$\text{Distance} = \frac{|\frac{7}{3} - 0|}{3} = \frac{\frac{7}{3}}{3} = \frac{7}{9}$$ **Final answer:** The distance between the two parallel planes is $$\boxed{\frac{7}{9}}$$ units.