1. In $\triangle A B C$, the three interior angles are $A, B, C$. If $\sin ^{2} A+\sin ^{2} B+\sin ^{2} C=2$, then the maximum value of $\cos A+\cos B+2 \cos C$ is
A. $\frac{3 \sqrt{2}}{2}$
B. $\sqrt{5}$
C. $\frac{4 \sqrt{3}}{3}$
D. $\sqrt{6}$
$$
\begin{array}{l}
\cos ^{2} A+\cos ^{2} B+\cos ^{2} C=1 \Rightarrow \frac{1}{2}(\cos 2 A+\cos 2 B)+\cos ^{2} C=0 \\
\Rightarrow-\cos C \cos (A-B)+\cos ^{2} C=-\cos C[\cos (A-B)+\cos (A+B)] \\
=-2 \cos A \cos B \cos C=0 \Rightarrow A, B, C \text { one of them is } 90^{\circ} .
\end{array}
$$

If one of $A, B$ is $90^{\circ}$, such as $A=90^{\circ}$, then $\cos A+\cos B+2 \cos C=\cos B+2 \sin B \leqslant \sqrt{5}$; if $C=90^{\circ}$, then $\cos A+\cos B+2 \cos C=\cos A+\sin A \leqslant \sqrt{2}$, so the answer is $B$.