D is the correct answer becoz ,
as we know that k5 is non -planner graph , it is prroved by corollary1 : if graph is planner with v veritices and e edges where v>=3, must have e<=3v-63v-6
as we know that k3-3 is non -planner graph , it is prroved by corollary 2: if graph is planner with v veritices and e edges where v>=3, must have e<=2v-4
now homemorphic and isoporphic both are different , so graph homeoporphic to any one of k5 or k3-3 then such graph is non-planner , but graph isomorphic to k5 and k3,3 is also non planner