In an equilateral triangle, the inradius, circumradius and one of the ex-radii are in the ratio

we have $\mathrm{\Delta }=\frac{\sqrt{3}}{4}{a}^2,s=\frac{3a}{2}$
$\begin{array}{r}\therefore r=\frac{\mathrm{\Delta }}{s}=\frac{a}{2\sqrt{3}},R=\frac{abc}{4\mathrm{\Delta }}=\frac{a^3}{\sqrt{3}{a}^2}=\frac{a}{\sqrt{3}}\\ \text{ and }{r}_1=\frac{\mathrm{\Delta }}{s-a}=\frac{\sqrt{3}/4a^2}{a/2}=\frac{\sqrt{3}}{2}a\end{array}$
$\text{ Hence, }\mathrm{r}:\mathrm{R}:{\mathrm{r}}_1=\frac{\mathrm{a}}{2\sqrt{3}}:\frac{\mathrm{a}}{\sqrt{3}}:\frac{\sqrt{3}}{2}\mathrm{a}=1:2:3$