若 $\tan \alpha=3$, 则 $\sin ^{2} \alpha-2 \sin \alpha \cos \alpha+3 \cos ^{2} \alpha=$
$\sin ^{2} \alpha-2 \sin \alpha \cos \alpha+3 \cos ^{2} \alpha$

$=\frac{\sin ^{2} \alpha-2 \sin \alpha \cos \alpha+3 \cos ^{2} \alpha}{\sin ^{2} \alpha+\cos ^{2} \alpha}$

$=\frac{\tan ^{2} \alpha-2 \tan \alpha+3}{\tan ^{2} \alpha+1}$

$=\frac{12-6}{10}=\frac{3}{5}$.