Explanation:1a. The dark line through the middle of the figure is intended to show where the planes intersect. Points A, D, and B are shown on the line. The line can be referred to using any two of those points: The line of intersection is line AB.__1b. The only three points shown on any line in the figure are the ones shown on the line of intersection: points A, B, D__1c. A plane can be named using any three non-collinear points on the plane. Possible names are ... plane ABF plane BDG__1d. The points shown to be in plane P are {A, B, C, D, E}. The points shown to be in plane R are {A, B, D, F, G}. To answer this question, you can choose any four points from either list. points A, B, F, G are coplanar__1e. In parts (a) and (b) we identified that points A, B, and D are on the line that lies in both planes. Any of these points will do. point A lies in both planes_____2. The distance formula can be used to find the distances between the points. Then those distances can be added to give the perimeter. AB = √((-1-(-2))^2+(4-1)^2) = √(1+9) = √10 ≈ 3.16 BC = √((2-(-2))^2+(1-1)^2) = √16 = 4 CA = √((2-(-1))^2+(1-4)^2) = √(9+9) = 3√2 ≈ 4.24So the perimeter is ... P = AB +BC +CA = 3.16 +4.00 +4.24 = 11.40The perimeter of ABC is about 11.4 units.