Answer: A, B, C
Loss of water gives CHX3−CX+(CX6HX5)−C(OH)(CX6HX5)CHX3, tertiary and benzylic. Phenyl has the higher migratory aptitude, so the product is CHX3−CO−C(CHX3)(CX6HX5)X2, 3,3-diphenylbutan-2-one. It contains CHX3COX− so it gives iodoform. The methyl-shift product, 2-methyl-1,2-diphenylpropan-1-one, is the minor one.