(i)     Consider the line from the common centre of $C_1$, $C_2$ and $C_3$ through the centre of one of the unit circles. This gives $r_1+2=r_3$. (ii)
Consider the equilateral triangle with vertices at the centres of the unit circles and side length $2$. The altitude of this triangle is therefore of length $\sqrt{3}$, which gives $1+r_1+r_2=\sqrt{3}$.

(iii)
The intersection of the altitudes of the triangle in (ii) divides each altitude in the ratio $2:1$, hence $2r_2=r_1+1$.

These three equations give

${\displaystyle r_1=\frac{2}{\sqrt{3}}-1,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{r}_2=\frac{1}{\sqrt{3}},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{r}_3=\frac{2}{\sqrt{3}}+1.}$