A Party of Three

From the case of three electrons,

$2\cfrac{qq}{4\pi\varepsilon_o{r}^{2}_{ee}} sin(60^o) = \cfrac{q.3q}{4\pi\varepsilon_o{r}^2_{pe}}$

$\cfrac{2}{{r}^{2}_{ee}}\cfrac{\sqrt{3}}{2} = \cfrac{3}{{r}^2_{pe}}$

$\sqrt{3}{r}^2_{ee}={r}^2_{pe}$

$\sqrt[4]{3}{r}_{ee}={r}_{pe}$

but from geometry,

$2.{r}_{pe}cos(30^o)={r}_{ee}$

$\sqrt{3}{r}_{pe}={r}_{ee}$

This means the configuration is not static, the electrons are in circular motion, instantaneously at least, in order not to fall into the positive charge.  Let assume orbit speed is $c$

$\cfrac{m_ec^2}{{r}_{pe}}= \cfrac{q.3q}{4\pi\varepsilon_o{r}^2_{pe}}-2\cfrac{qq}{4\pi\varepsilon_o{r}^{2}_{ee}} sin(60^o)$

$\cfrac{m_ec^2}{{r}_{pe}}= \cfrac{3q^2}{4\pi\varepsilon_o{r}^2_{pe}}-2\cfrac{q^2}{4\pi\varepsilon_o 3{r}^{2}_{pe}} .\cfrac{\sqrt{3}}{2}$

${{r}_{pe}}= \cfrac{q^2}{4\pi\varepsilon_o m_ec^2}(3-\cfrac{\sqrt{3}}{3})$

${{r}_{pe}}$ = (1.60217657e-19)^2/(4*pi*8.8541878176e-12*9.10938291e-31*299792458^2)*(3-(3)^(0.5)/3)

${{r}_{pe}}$ = 6.8269e-15 m

But the atomic radius of Lithium. Li is 1.52e-10 m.  There is one possibility, that the electrons are not orbiting a light speed $c$.  What speed are the electron orbiting at then?  Then again electron mass has a relative uncertainty of 4.2e-10, so what of e-15.  This value can not be more accurate than electron mass.