11. Let the three interior angles $A, B, C$ of $\triangle ABC$ form an arithmetic sequence, and the reciprocals of the three side lengths $a, b, c$ also form an arithmetic sequence. Try to find $A, B, C$.
11. Let $A=B-\alpha, C=B+\alpha(90>\alpha \geqslant 0), A+B+C=B-\alpha+\beta+B+\alpha=180^{\circ} \Rightarrow$ $B=60^{\circ}$. Also, $\frac{1}{a}+\frac{1}{c}=\frac{2}{b} \Leftrightarrow b c+c a=2 a c \Leftrightarrow 2 R \sin B \cdot 2 R \sin C+2 R \sin C \cdot 2 R \sin A=2$. $2 R \sin A \cdot 2 R \sin C \Leftrightarrow \sin B \sin C+\sin B \sin A=2 \sin A \sin C \Leftrightarrow \frac{3}{2} \cos \alpha=2 \cos ^{2} \alpha-\frac{1}{2}, \cos \alpha=1$ or $\cos \alpha=-\frac{1}{4}$. Since $0 \leqslant \alpha<90^{\circ} \Rightarrow \cos \alpha \neq-\frac{1}{4}$, thus $\cos \alpha=1 \Rightarrow \alpha=0$, hence $A=B=C=60^{\circ}$.