Therefore, a(s) = 1 + (−3)2 + (−2)2 da d = 14 da. d since we want the area of the section of the plane that lies above the first octant, then the region d will be the triangular region bounded by the x-axis, the y-axis, and the line formed by the intersection of the plane with the plane z = 0. substituting z = 0 into 3x + 2y + z = 6 and solving for y, we find that this line of intersection has the equation