Give $3$ spheres in Cartesian coordinates (I write them in vector form because otherwise they will be long) $$\begin{cases} \Vert\mathbf{x}-\mathbf{x}_0\Vert=r\\ \Vert\mathbf{x}-\mathbf{x}_1\Vert=r_1\\ \Vert\mathbf{x}-\mathbf{x}_2\Vert=r_2\end{cases}$$
The intersection point can be calculated like this (I calculated the formula myself):
$$\mathbf{v}_1:=\mathbf{x}_1-\mathbf{x}_0\qquad \mathbf{v}_2:=\mathbf{x}_2-\mathbf{x}_0$$ $$\mathbf{S}:=\mathbf{v}_1\times\mathbf{v}_2$$ $$\mathbf{T}:=\frac{1}{2}\begin{vmatrix} \mathbf{v}_1&\Vert\mathbf{v}_1\Vert^2-r_1^2+r^2\\ \mathbf{v}_2&\Vert\mathbf{v}_2\Vert^2-r_2^2+r^2 \end{vmatrix}$$
$$\mathbf{x}=\mathbf{x}_0+\hat{\mathbf{S}}\times\frac{\mathbf{T}}{\Vert\mathbf{S}\Vert}\pm\hat{\mathbf{S}}\sqrt{r^2-\left(\frac{\Vert\mathbf{T}\Vert}{\Vert\mathbf{S}\Vert}\right)^2}$$
Would it be possible to rewrite this formula without making it depend so directly only on the first sphere? I would like to obtain a "symmetric" formula.
For example $\Vert\mathbf{S}\Vert$, is twice the area of the triangle which has the centers as vertices and does not depend "disproportionately" on any of the data
Are there more "smart" points to consider? Like the barycenter of the triangle, the orthocenter, the circumcenter etc...?
Comment: The intersection of each two spheres is a circle,So we have three circles a, b and c with three equations like:
$(x-m)^2+(y-n)^2+(z-s)^2=R^2\space\space\space\space\space\space\space\space(1)$
where m, n and t represent coordinates of the centers of circles a, b and c and R the radius.Equations (1) is the solution of system of equations such as:
$\begin{cases}Sphere 1:(x-x_1)^2+(y-y_1)^2+(z-z_1)^2=r_1^2\\Sphere 2:(x-x_2)^2+(y-y_2)^2+(z-z_2)^2=r_2^2\end{cases}\rightarrow (x-m)^2+(y-n)^2+(z-s)^2=R^2 $
So we have three systems of equations with equations of spheres (1&2), (2&3) and (1&3). As can be seen in picture 2 the final result is the coordinates in 3D of two points:
For simplicity you may consider origin as the center one sphere.