One thing that may be causing confusion is that the problem is not stated very well. We are told the figure is bounded by the three coordinate planes, by the plane x+ y+ z= 2, and by the plane z= x+ y. But there is NO region bounded by those five planes! The plane z= x+ y "covers" the plane z= 0. I think what is intended is the region bounded by the two coordinate planes, x= 0 and y= 0, and by x+ y+ z= 2 and z= x+ y but not by the plane z= 0. Adding x+ y+ z= 2 and x+ y- z= 0 we get 2x+ 2y= 2 or x+ y= 1. The last two planes intersect at the line x+ y= 1, z= 1. The first plane, x+ y+ z, intersects the coordinate planes in a triangle with vertices (0, 2, 0), (2, 0, 0) and (0, 0, 2). The sides have equations x= 0, y+ z= 2, y= 0, x+ z= 2, and z= 0, x+ y= 2.
The second triangle passes through the origin, (0, 0, 0), and cuts the first plane in the line x+ y= 1, z= 1.
The largest value x take in the region is 1 so we must take x from 0 to 1. For every x, y goes from 0 to 1- x, and for every x, y, z goes from z= 2- x- y.
The integral can be written a $\int_{x= 0}^1\int_{y= 0}^{1- x}\int_{z= 0}^{2- x- y}dzdxdy$.