Correct option is A213−12113+13+13+13\frac{{2\frac{1}{3}} - 1\frac{2}{11}}{3 + \frac{1}{3 + \frac{1}{3 + \frac{1}{3}}}}3+3+3+3111231−1112=73−13113+13+13+13\frac{{\frac{7}{3}} - \frac{13}{11}}{3 + \frac{1}{3 + \frac{1}{3 + \frac{1}{3}}}}3+3+3+311137−1113=73−1311333+1033\frac{{\frac{7}{3}} - \frac{13}{11}}{ \frac{33}{3 + \frac{10}{33}}}3+33103337−1113=3833×33109\frac{38}{33}\times\frac{33}{109}3338×10933=38109\frac{38}{109}10938Hence answer is38109\frac{38}{109}10938