Ratio of angles is given by $\angle \mathrm{A}: \angle \mathrm{B}: \angle \mathrm{C}=1: 2: 3$
Let $\angle \mathrm{A}=\mathrm{x}, \angle \mathrm{B}=2 \mathrm{x}$ and $\angle \mathrm{C}=3 \mathrm{x}$


We know that in a triangle $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}$
$\Rightarrow \mathrm{x}+2 \mathrm{x}+3 \mathrm{x}=180^{\circ}$
$\Rightarrow \quad x=\frac{180^{\circ}}{6}=30^{\circ}$
So, $\angle \mathrm{A}=30^{\circ}, \angle \mathrm{B}=60^{\circ}$ and $\angle \mathrm{C}=90^{\circ}$
From sin law
$\frac{\mathrm{BC}}{\sin \mathrm{A}}=\frac{\mathrm{CA}}{\sin \mathrm{B}}=\frac{\mathrm{AB}}{\sin \mathrm{C}}=\mathrm{K}$
$\frac{\mathrm{BC}}{\sin 30^{\circ}}=\frac{\mathrm{CA}}{\sin 60^{\circ}}=\frac{\mathrm{AB}}{\sin 90^{\circ}}=\mathrm{K}$
$\mathrm{BC}=\mathrm{K} \sin 30^{\circ}$
$\mathrm{CA}=\mathrm{K} \sin 60^{\circ}$
$\mathrm{AB}=\mathrm{K} \sin 90^{\circ}$
$\mathrm{BC}: \mathrm{CA}: \mathrm{AB}$
$=\sin 30^{\circ}: \sin 60^{\circ}: \sin 90^{\circ}=\frac{1}{2}: \frac{\sqrt{3}}{2}: 1=1: \sqrt{3}: 2$