I thought to post this doubt in mathematics section as I thought It was more of a vector question. I haven't studied vectors but there is some use in physics. Is there any theorem from which I can get the value of the above mentioned equation in the image. I was willing to find multiple equillibrium positions rather than finding only centroid of the polygon as an equllibrium position. I am sorry if I had done something wrong in anyway , I am too naive in this.
As per my comment, you don't need for the sum of inverse squared magnitudes to be zero as it will never be zero (as Gerry Myerson pointed out). What you need is for the vector sum to be zero so
$$ \sum_\ell \frac{\hat{\mathbf{r}}_\ell}{\| \mathbf{r}_\ell \|^2} = \vec{0} $$
where $\hat{\mathbf{r}}_\ell$ are the unit vectors pointing from your "test point" to particle $\ell$ in your charge distribution (or vice versa).
We can try to think of it geometrically but a closed form solution will be hard to come by on pure geometrical intuition. "Intuition" can also be misleading. I can think of at least $N+1$ points where I would imagine that the forces might sum to zero but I cannot guarantee this on intuition alone. As it turns out, analysis shows that only one of these points comes to equilibrium.
How do we know if there is only one or if there are more? We either have to have strokes of geometric genius or we can convert the problem into an algebraic one.
Consider a distribution with $N$ same-magnitude charges on the corners of a regular n-polygon. We can draw a circle such that it includes all the charges and they will be separated by equal arcs. We call this the circumscribed circle. We will develop a system that solves for the points where the electric field is equal to zero.
Since each charge in the distribution is located on a circle, we can say that charge $\ell$ is at
$$ \theta_\ell = \frac{2\pi(\ell - 1)}{N} \: \text{ where } \: k=1,2,\dots,N \text{ .}$$
Now we can say that the cartesian coordinate of charge $\ell$ is $(x,y) = (R\cos\theta_\ell,R\sin\theta_\ell)$ where $R$ is the radius of the circumscribing circle (see appendicized argument if you don't know what this is). We will now adjust our expression for the electric field as
$$ \vec{\mathbf{E}}(x,y) = kQ\sum_\ell \frac{\hat{\mathbf{r}}_\ell}{\| \mathbf{r}_\ell \|^2} = kQ\sum_\ell \frac{\vec{\mathbf{r}}_\ell}{\| \mathbf{r}_\ell \|^3} $$
$$\text{where} $$
$$ \vec{\mathbf{r}}_\ell = \begin{bmatrix} R\cos\theta_\ell - x\\ R\sin\theta_\ell - y \end{bmatrix}$$
$$\text{and}$$
$$\| \mathbf{r}_\ell \|^2 = (x - R\cos\theta_\ell)^2 + (y - R\sin\theta_\ell)^2 \text{ .}$$
We then have the following non-linear system of equations for the equilibria
$$ \vec{\mathbf{E}}(x,y) = kQ\sum_\ell \bigg( (x - R\cos\theta_\ell)^2 + (y - R\sin\theta_\ell)^2 \bigg)^{-3/2} \begin{bmatrix} R\cos\theta_\ell - x\\ R\sin\theta_\ell - y \end{bmatrix} = \begin{bmatrix} 0\\0 \end{bmatrix} \text{ .}$$
A solution to the above equations would yield all equilibria. I personally think it is best left to a computational solver.
Let me know if you have any questions or comments or if I've made any glaring errors.