I need to determine the interception location of three spheres when they are in different locations besides the origin. I feel that this is pretty straight forward, but I can't seem to get the algebra correct when combining equations. I need to find the interception point $(x,y,z)$ of three spheres. I've been banging my head on this problem for several days now, so any help will be appreciated.
The equations of the spheres are...

*

*$R^2=(x-a)^2+(y-b)^2+(z-c)^2$

*$S^2=(x-g)^2+(y-h)^2+(z-j)^2$

*$T^2=(x-n)^2+(y-o)^2+(z-p)^2$
All variables are known except the intercept location of (x,y,z). I'm writing an algorithm that will calculate this in excel, so ideally I'd prefer an equation for each x,y,z coordinate.
Thanks in advance.
 A: The given equations of the three spheres are
$R^2 = (x - a)^2 + (y - b)^2 + (z - c)^2\\
S^2 = (x - g)^2 + (y - h)^2 + (z - j)^2 \\
T^2 = (x - n)^2 + (y - o)^2 + (z - p)^2$
Subtracting the second equation from the first, results in
$ K_1 = -2 x (a - g) - 2 y (b - h) - 2 z(c - j) \hspace{20pt}(1)$
where
$K_1 = R^2 - S^2 + g^2 + h^2 + j^2 - a^2 - b^2 - c^2$
And subtracting the third equation from the first equation gives us
$ K_2 = -2 x(a - n) - 2 y (b - o) - 2 z (c - p)\hspace{20pt}(2) $
where
$ K_2 = R^2 - T^2 + n^2 + o^2 + p^2 - a^2 - b^2 - c^2$
Equations $(1)$ and $(2)$ are equations of planes and can be solved simultaneously using Gauss-Jordan elimination, and this will result in
$ (x, y, z) = (x_0, y_0, z_0) + t (v_1, v_2, v_3)\hspace{20pt} (3) $
where $x_0, y_0, z_0, v_1, v_2, v_3$ have known numeric values, and $t$ is a free parameter, i.e. $t \in \mathbb{R} $
Substituting (3) into one of the original equations of the spheres, we can solve for $t$ (there can be 2 values or 1 value or no solutions).  Once we have the values of $t$ we substitute that into $(3)$ to obtain the value of $(x,y,z)$.
