In the context of determining pore volumes in adsorbing materials, I'm trying to find the pores that a gas molecule can go through. To do so, I need to solve a geometry/linear algebra problem. Here is a figure representing the problem. 
There are three circles with center coordinates C1, C2, and C3 with a radius r1, r2, r3, and I would like to calculate the radius of the circle, R, that is inside and touching the three circles at coordinate C. Note that $C, C_1, C_2 , C_3 \in \mathbb{R}^3$
I have already gotten a great answer for a similar case. but it was for spheres in 3D, not circles in 3D. Based on this answer solution for 2D case can be derived.
$A = \begin{bmatrix} (C_1 - C_2)^T \\ (C_1 - C_3)^T \end{bmatrix}$
$b = \begin{bmatrix} r_2 - r_1 \\ r_3 - r_1 \end{bmatrix}$
$c = \dfrac{1}{2} \begin{bmatrix} r_2^2 - r_1^2 - \| C_1 - C_2 \|^2 \\ r_3^2 - r_1^2 - \| C_1 - C_3 \|^2 \end{bmatrix}$
And we can solve the quadratic equation to get R: $R^2 + 2 R r_1 + r_1^2 = R^2 \bigg( b^T A^{-T}A^{-1} b \bigg) + 2 R \bigg( b^T A^{-T} A^{-1} c \bigg) + \bigg( c^T A^{-T} A^{-1} c \bigg)$
But this equation is only for 2D and not for 3D. If I apply this equation for the 3D case, $A \in \mathbb{R}^{2\times3}$, so I cannot calculate $A^{-1}$ as it is not a square matrix. I could rotate the plane so that values in one axis is 0, but I'm wondering if there is a more concise calculation.
We could rotate the triangle to a plane normal to an axis, and drop the dimension of coordinates corresponding to the axis, but this yield no solution if the triangle plane is already normal to the axis. This is problematic when trying to vectorize for fast computation.
The solution I found is from Invertible matrix of non-square matrix?.
Where we can convert the A to a square matrix by:
$A^{-1}=A^\top(AA^\top)^{-1}$
By applying this equation, we can get the R for 2D triangle in 3D.