1. Given that the three interior angles $\angle A, \angle B, \angle C$ of $\triangle \mathrm{ABC}$ satisfy $2 \angle A - \angle B, 3 \angle B - 2 \angle C$, and $90^{\circ} - \angle A$, the minimum of these is $\alpha$, then the maximum value of $\alpha$ is $\qquad$
1. $\frac{2 \pi}{9}$

Analysis: Using the method of undetermined coefficients and the maximum value inequality, from the given conditions we can obtain:
$$
\begin{array}{l}
\alpha=\min \left\{2 \angle A-\angle B, 3 \angle B-2 \angle C, 90^{\circ}-\angle A\right\} \leq \frac{\lambda(2 \angle A-\angle B)+\mu(3 \angle B-2 \angle C)+\left(90^{\circ}-\angle A\right)}{\lambda+\mu+1} \\
\frac{\left(2 \lambda-\frac{1}{2}\right) \cdot \angle A+\left(3 \mu-\lambda+\frac{1}{2}\right) \cdot \angle B+\left(\frac{1}{2}-2 \mu\right) \cdot \angle C}{\lambda+\mu+1} \text { holds for any } \lambda, \mu \in R \text { . When } \\
\lambda=\frac{5}{12}, \mu=\frac{1}{12} \text { satisfy } 2 \lambda-\frac{1}{2}=3 \mu-\lambda+\frac{1}{2}=\frac{1}{2}-2 \mu=\frac{1}{3}, \text { at this time } \alpha \leq \frac{2 \pi}{9}
\end{array}
$$