In 2-space, this would be an evaluation question. It follows the process in which you choose a value of a: to evaluate at, and then solve for the value of y. In 3-space with 3 variables, you choose values for 2 of the variables and solve for the remaining one. Lets=1,y=1, II o: -> 2(1) (1} + fit2+1 II on This means that the point (1, 1, 5) is on the plane 23: y + 32: = 6. Note The easiest points to nd are the s, y, and s intercepts. To solve for an intercept, you set the other two variables equal to zero. While more difcult, the choice of a: = 1, y = 1 better demonstrates the process