To solve this problem, we need to find the points on the intersection of the plane $x + y + 2z = 2$ and the paraboloid $z = x^2 + y^2$ that are nearest to and farthest from the origin. Here’s a step-by-step approach to the solution:

Step 1: Substitute $z = x^2 + y^2$ into the Plane Equation

Since $z = x^2 + y^2$ on the paraboloid, substitute this into the plane equation: $x + y + 2(x^2 + y^2) = 2$ This simplifies to: [ x + y + 2x^2 +