<s>[INST] 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$=43$
 [/INST] As @Unit commented, Herons formula might be an easier bet, we have:
$$\overrightarrow{AB}=\begin{pmatrix}3-1\\-6-2\\-4+3\\2-3\end{pmatrix}=\begin{pmatrix}2\\-8\\-1\\-1\end{pmatrix}\Rightarrow\left|\overrightarrow{AB}\right|=\sqrt{2^2+8^2+1^2+1^2}=\sqrt{70}$$
$$\overrightarrow{BC}=\begin{pmatrix}-6\\-10\\0\\-2\end{pmatrix}\Rightarrow\left|\overrightarrow{BC}\right|=\sqrt{6^2+10^2+2^2}=\sqrt{140}$$
$$\overrightarrow{AC}=\begin{pmatrix}-4\\-18\\-1\\-3\end{pmatrix}\Rightarrow\left|\overrightarrow{AC}\right|=\sqrt{4^2+18^2+1^2+3^2}=\sqrt{350}$$

We can say that the Area of the triangle is:
$$A_{abc}=\sqrt{S\left(S-|\overrightarrow{AC}|\right)\left(S-|\overrightarrow{BC}|\right)\left(S-|\overrightarrow{AB}|\right)}$$
where:
$$S=\frac{|\overrightarrow{AC}|+|\overrightarrow{AB}|+|\overrightarrow{BC}|}{2}$$
</s>