smc

Let the center of the $3$" circle be $O$, that of the $2$" circle be $A$, that of the $1$" circle be $B$, and those of the circles of unknown radius (let their radii have length $r$) be $C$ and $D$, as in the diagram. Also, in this problem, no two circles share a center, so let the circles be named by their corresponding centers (so, the $3$" circle is circle $O$, etc.). Extend $\overline{OC}$ past $C$ to intersect circle $O$ at point $E$. Because circle $O$ has radius $\frac{3}{2}$ and circle $A$ has radius $1$, $OA=\frac{3}{2}-1=\frac{1}{2}$. Likewise, because $B$ has radius $\frac{1}{2}$, $OB=\frac{3}{2}-\frac{1}{2}=1$. Thus, $AB=\frac{3}{2}$. Furthermore, because the line connecting the centers of two tangent circles goes through their point of tangency, $AC=r+1$ and $BC=r+\frac{1}{2}$. Because $OE=\frac{3}{2}$ and $CE=r$, $OC=\frac{3}{2}-r$. With this information, we can now apply Stewart's Theorem to $△ABC$ to solve for $r$: $$\begin{array}{rl}OB\ast OA\ast AB+O{C}^2\ast AB& =A{C}^2\ast OB+B{C}^2\ast OA\\ 4[(1)(\frac{1}{2})(\frac{3}{2})+(\frac{3}{2}-r{)}^2(\frac{3}{2})]& =4[(r+1{)}^2(1)+(r+\frac{1}{2}{)}^2(\frac{1}{2})]\\ 3+6(\frac{9}{4}-3r+r^2)& =4(r^2+2r+1)+2(r^2+r+\frac{1}{4})\\ 3+\frac{27}{2}-18r+6r^2& =4r^2+8r+4+2r^2+2r+\frac{1}{2}\\ 3+\frac{27-1}{2}-4+6r^2-6r^2& =18r+8r+2r\\ 3+13-4& =28r\\ 28r& =12\\ r& =\frac{3}{7}\end{array}$$ Because the question asks for the diameter of the circle, we calculate $2r=\frac{6}{7}\approx \boxed{.86}$.