Looking for the simplest method to calculate the centre point between three 3-Dimensional coordinates. I've spent a lot of time googling around and I don't think I ever did the kind of math I saw, and several people mentioned converting to 2 dimensions and then back but I don't understand how that doesn't affect the fidelity of the 3rd dimension and although it appears a few people have built the necessary libraries for a calculator (something called a miniball, if I understood it correctly), no one seems to have actually implemented one in a convenient format. The one online calculator I found that claimed it would do what I needed also required a fourth point, which I don't have.
So, three points: (-779,-51-327), (-767,-52,-279), and (-731,-36,-283), from which I need the point in the centre of the three.
I made it to Calculus 1 in college, but that was a really long time ago and while I'm happy to do the math by hand if necessary, I'll need to know the names of the operations if I'm going to look up what they are (something which was missing from the other examples I mentioned). Thanks.
If by "centre" you mean centroid or barycentre – the intersection of the medians of the triangle –, then $$\left\lbrace ~ \begin{aligned} x &= \frac{-779 ~+~ -767 ~+~ -731}{3} = \frac{-2277}{3} = -759 \\ y &= \frac{-51 ~+~ -52 ~+~ -36}{3} = \frac{-139}{3} \approx -46.333 \\ z &= \frac{-327 ~+~ -279 ~+~ -283}{3} = \frac{-889}{3} \approx -296.333 \\ \end{aligned} \right .$$
Other possible "centres" are orthocenter (the intersection of the altitudes of the triangle), circumcenter (the center of the circle passing through the three points; also the intersection of the triangle angle bisectors), incenter (the center of the circle inscribed within the triangle), and the center of the nine-point circle.
In those other cases, the simple solution is indeed to change to a 2D coordinate system with one of the points at 2D origin, one of the points at $(a, 0)$ on the positive $x$ axis, and the third point at $(b, c)$ (typically choosing $c \gt 0$). Numerically, you'll want to label the points so that the longest edge is between the first two vertices.
In this case, we'll choose $\vec{p}_1 = (-779, -51, -327)$, $\vec{p}_2 = (-731, -36, -283)$, and $\vec{p}_3 = (-767, -52, -279)$. $\vec{p}_1$ will be at 2D origin, $\vec{p}_2$ at $(a, 0)$, and $\vec{p}_3$ at $(b, c)$. $$a = \left\lVert \vec{p}_2 - \vec{p}_1 \right\rVert = \sqrt{ (-731 - -779)^2 + (-36 - -51)^2 + (-283 - -327)^2 } = \sqrt{ (48)^2 + (15)^2 + (44)^2 } = \sqrt{4465} \approx 66.820655$$ This leaves the third point, $\vec{p}_3$. We know it is at distances $$\begin{aligned} d_{13} &= \left\lVert \vec{p}_3 - \vec{p}_1 \right\rVert = \sqrt{ 2449 } \approx 49.487372 \\ d_{23} &= \left\lVert \vec{p}_3 - \vec{p}_2 \right\rVert = \sqrt{ 1568 } \approx 39.597980 \\ \end{aligned}$$ from the two vertices. Since the triangle orientation does not matter, we can arbitrarily choose $c \gt 0$. As the 2D edge lengths of the triangle must match, we have $$\left\lbrace ~ \begin{aligned} d_{13}^2 &= b^2 + c^2 \\ d_{23}^2 &= (a - b)^2 + c^2 = a^2 + b^2 + c^2 - 2 a b \\ \end{aligned} \right .$$ which we can solve for $b$ and $c \gt 0$: $$\left\lbrace ~ \begin{aligned} b &= \displaystyle \frac{a}{2} + \frac{d_{13}^2 - d_{23}^2}{2 a} \\ c &= \displaystyle \frac{1}{2 a} \sqrt{ (a + d_{13} + d_{23}) (a + d_{13} - d_{23}) (a - d_{13} + d_{23}) (-a + d_{13} + d_{23}) } \\ \end{aligned} \right .$$ Note that this applies to all triplets of 3D points that are not collinear, and it is numerically most stable (including smallest errors when using floating-point calculations) when $a \ge d_{13}$ and $a \ge d_{23}$.
Then, you use the known Cartesian solutions for the "centre" you want, noting that the triangle vertices are $(0, 0)$, $(a, 0)$, and $(b, c)$.
Note that $c \gt 0$, because otherwise the three points would be collinear (and we already chose $c$ to be positive, as the orientation of the triangle [clockwise or counterclockwise in 2D] does not matter here), and that $a \ne 0$, because otherwise the first two points would be indistinguishable (and the triangle degenerate, a line segment or a point).
(If you did choose the longest edge to be between the first two vertices, then $a \ge \lvert b \rvert$, and $a \ge c \gt 0$.)
Let's say the coordinates of the "centre" are $(u, v)$ in these coordinates.
We can easily convert back to 3D coordinates using barycentric coordinates $[w_2, w_3, w_1]$, since $$\left\lbrace ~ \begin{aligned} w_2 a + w_3 b &= u \\ w_3 c &= v \\ w_1 &= 1 - w_2 - w_3 \\ \end{aligned} \right .$$ the solution of which is $$\left\lbrace ~ \begin{aligned} w_3 &= \displaystyle \frac{v}{c} \\ w_2 &= \displaystyle \frac{c u - b v}{a c} \\ w_1 &= 1 - w_2 - w_3 \\ \end{aligned} \right .$$
The 3D location of the point is $$\vec{p} = w_1 \vec{p}_1 + w_2 \vec{p}_2 + w_3 \vec{p}_3$$