Given a triangle in $\Bbb R^3$ with vertices $A = (x_1,y_1,z_1)$, $B = (x_2,y_2,z_2)$, and $C = (x_3,y_3,z_3)$ the formula to find the center of gravity $G$ is
$$G= \left(\frac{1}{3}\displaystyle\sum_{i=1}^3x_i, \frac{1}{3}\displaystyle\sum_{i=1}^3y_i, \frac{1}{3}\displaystyle\sum_{i=1}^3z_i \right) \tag 1$$
where you simply take the component-wise average of the vertices. For my 16-year-old students I prove formula $(1)$ in $\Bbb R^2$ using the fact that
Each median remains divided from the centre of gravity into two parts and the one to which the vertex belongs is double of the other.
What other methods are there to compute the center of gravity of a triangle in $\Bbb{R}^3$? Are there methods that don't use goniometry, or don't even consider the center as the intersection of the three medians? You could obviously compute it with a double-integral. Is there a way to compute this using a determinant or other methods of linear algebra?