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
$$\vec{p} = w_1 \vec{p}_1 + w_2 \vec{p}_2 + w_3 \vec{p}_3$$
