1. **State the problem:** Calculate the triple integral $$\int_0^3 \int_0^2 \int_0^{3-2x} dz\,dy\,dx$$. 2. **Understand the integral:** The integral is over $z$, $y$, and $x$ with limits: - $x$ from 0 to 3, - $y$ from 0 to 2, - $z$ from 0 to $3 - 2x$. 3. **Integrate with respect to $z$ first:** Since the integrand is 1 (no function given), the integral over $z$ is: $$\int_0^{3-2x} 1 \, dz = z \Big|_0^{3-2x} = 3 - 2x$$ 4. **Substitute back and integrate over $y$:** $$\int_0^2 (3 - 2x) \, dy = (3 - 2x) y \Big|_0^2 = 2(3 - 2x) = 6 - 4x$$ 5. **Integrate over $x$:** $$\int_0^3 (6 - 4x) \, dx = \int_0^3 6 \, dx - \int_0^3 4x \, dx$$ 6. **Calculate each integral:** $$6x \Big|_0^3 = 6 \times 3 = 18$$ $$2x^2 \Big|_0^3 = 2 \times 9 = 18$$ 7. **Combine results:** $$18 - 18 = 0$$ 8. **Check intermediate cancellation:** When integrating $\int_0^3 (6 - 4x) dx$, write as: $$\int_0^3 \cancel{2}(3 - 2x) dx$$ **Final answer:** $$\boxed{0}$$