Correct answer: Qg21.Qc8? (2.Qc2#) 1...K:b3/[б]/ 2.Qb7#[A] 1...a:b3[c] 2.Qc1#[Б] but 1...R:b3+[a]! 1.Bb4[C]? zz 1...R:b3[a]/a:b3[c] 2.Qg2#[E] 1...K:b3/[б
