Area of a Triangle in $\mathbb{R}^4$ I find this question quite tricky, and I don't know if this kind of treatment is right.
I was asked about the area of a triangle formed by the points: $A:(1,2,-3,3)$; $B:(3,-6,-4,2)$; and $C:(-3,-16,-4,0)$. The only way I could make a reason out of this would be getting all the $2\times2$ matrices within the matrix M whose columns are AB and AC, getting their determinants, and finally dividing it by two. Is this procedure right? I will write the processes below:
Firstly, get vectors AB and AC:

$AB = (2,-8,-1,-1)$

$AC = (-4,-18,-1,-3)$
Secondly, construct matrix M:
\begin{equation}
M_{4,2}=
\begin{pmatrix}
2 & -4  \\
-8 & -18  \\
-1  & -1  \\
-1 & -3 
\end{pmatrix}
\end{equation}
Thirdly, get all the $2\times2$ determinants within M:
\begin{equation}
S_{2}=
\begin{vmatrix}
2 & -4 \\ 
-8 & -18
\end{vmatrix}
+
\begin{vmatrix}
2 & -4 \\ 
-1 & -1
\end{vmatrix}
+
\begin{vmatrix}
2 & -4 \\ 
-1 & -3
\end{vmatrix}
+
\begin{vmatrix}
-8 & -18 \\ 
-1 & -1
\end{vmatrix}
+
\begin{vmatrix}
-8 & -18 \\ 
-1 & -3
\end{vmatrix}
+
\begin{vmatrix}
-1 & -1 \\ 
-1 & -3
\end{vmatrix}
\end{equation}
Forthly, dividing by two, and getting the absolute value (Area=$|\frac{S_{2}}{2}|$):
Area$=43$

Given $n$-vectors $x_1,\dots,x_n\in\mathbb R^m$, the $n$-dimensional volume of the parallelepiped spanned by $x_1,\dots,x_n$ is given by $\sqrt{\det(G)}$, where $G$ is the Gramian matrix given by $G=(\langle x_i, x_j\rangle)_{ij}$.
Hence, we get
$$
A = \frac{\sqrt{\det(X^T X)}}{2}=35,\quad \text{where}\quad X=\begin{pmatrix}2&-4\\-8&-18\\-1&-1\\-1&-3\end{pmatrix}.
$$
