1. **State the problem:** We are given the surface defined by the equation $$z = -2y^2 - x^2 + 54$$ and two planes $$x = 3$$ and $$y = 3$$. 2. **Find the intersection with the plane $$x=3$$:** Substitute $$x=3$$ into the surface equation: $$z = -2y^2 - (3)^2 + 54 = -2y^2 - 9 + 54 = -2y^2 + 45$$ This represents a parabola in the $$yz$$-plane. 3. **Find the intersection with the plane $$y=3$$:** Substitute $$y=3$$ into the surface equation: $$z = -2(3)^2 - x^2 + 54 = -18 - x^2 + 54 = -x^2 + 36$$ This represents a parabola in the $$xz$$-plane. 4. **Summary:** - On the plane $$x=3$$, the curve is $$z = -2y^2 + 45$$. - On the plane $$y=3$$, the curve is $$z = -x^2 + 36$$. These are the cross sections of the paraboloid with the given planes. **Final answers:** $$\text{At } x=3: z = -2y^2 + 45$$ $$\text{At } y=3: z = -x^2 + 36$$