Correct answer: Qa11...Ke6[a] 2.Qb3# 1...Kc6/[б]/ 2.Qh1#[A]/Qf3# 1.Qd2? zz 1...Ke6[a] 2.Qa2#[Б] 1...Kc6/[б]/ 2.Qg2#[C] but 1...K:d4! 1.Qc1? zz