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$=43$
 A: 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
$$
A = \frac{\sqrt{\det(X^T X)}}{2}=35,\quad \text{where}\quad X=\begin{pmatrix}2&-4\\-8&-18\\-1&-1\\-1&-3\end{pmatrix}.
$$
A: 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:
$$S=\frac{|\overrightarrow{AC}|+|\overrightarrow{AB}|+|\overrightarrow{BC}|}{2}$$
A: In the  $m$-D space , we generalize the cross product by the wedge product.
Given two vectors ${\bf a} = \vec {AB} , \; {\bf b} = \vec {AC} $, expressed in the base system
$$
\eqalign{
  & {\bf a} = a_{\,1} {\bf e}_{\,1}  + a_{\,2} {\bf e}_{\,2}  +  \cdots  + a_{\,m} {\bf e}_{\,m}   \cr 
  & {\bf b} = b_{\,1} {\bf e}_{\,1}  + b_{\,2} {\bf e}_{\,2}  +  \cdots  + b_{\,m} {\bf e}_{\,m}  \cr} 
$$
their wedge product is defined to be the 2-vector (or 2-blade)
$$
\eqalign{
  & {\bf w} = {\bf a} \wedge {\bf b} =   \cr 
  &  = \left( {a_{\,1} {\bf e}_{\,1}  + a_{\,2} {\bf e}_{\,2}  +  \cdots  + a_{\,m} {\bf e}_{\,m} } \right)
 \wedge \left( {b_{\,1} {\bf e}_{\,1}  + b_{\,2} {\bf e}_{\,2}  +  \cdots  + b_{\,m} {\bf e}_{\,m} } \right) =   \cr 
  &  = \sum\limits_{1\, \le \,k,j\, \le m} {a_{\,k} b_{\,j} {\bf e}_{\,k}  \wedge {\bf e}_{\,j} }
  = \sum\limits_{1\, \le \,k < j\, \le m} {\left( {a_{\,k} b_{\,j}  - a_{\,j} b_{\,k} } \right){\bf e}_{\,k}  \wedge {\bf e}_{\,j} }  \cr} 
$$
a total of $\binom {m}{2}$ base components since, by definition
$$
{\bf e}_{\,k}  \wedge {\bf e}_{\,j}  =  - \,{\bf e}_{\,j}  \wedge {\bf e}_{\,k}
 \quad {\bf e}_{\,k}  \wedge {\bf e}_{\,k}  = 0
$$
The magnitude of $\bf w$ is defined as
$$
\left| {\bf w} \right| = \left| {{\bf a} \wedge {\bf b}} \right|
 = \sqrt {\sum\limits_{1\, \le \,k < j\, \le m}
 {\left( {a_{\,k} b_{\,j}  - a_{\,j} b_{\,k} } \right)^{\,2} } } 
$$
So, for $m=2$ we get
$$
\left| {\bf w} \right| = \left| {a_{\,1} b_{\,2}  - a_{\,2} b_{\,1} } \right|
$$
which is the area of the parallelogram with sides defined by ${\bf a}, {\bf b}$,
and for $m=3$
$$
\left| {\bf w} \right|
 = \sqrt {\left| {a_{\,1} b_{\,2}  - a_{\,2} b_{\,1} } \right|^{\,2} 
 + \left| {a_{\,1} b_{\,3}  - a_{\,3} b_{\,1} } \right|^{\,2} 
 + \left| {a_{\,2} b_{\,3}  - a_{\,3} b_{\,2} } \right|^{\,2} } 
$$
which corresponds to the modulus of the cross-product ${\bf a} \times {\bf b}$,
and therefore again to the area of the parallelogram defined by by ${\bf a}, {\bf b}$.
And the above generalizes to ${\mathbb R}^m$ where the magnitude of the wedge product
of two vectors equal the area of the 2D parallelotope defined by the two vectors, and where
the magnitude of each component represents the area of the projection of the parallelotope onto each coordinate
plane.
But for what concerns the modulus, we can also express it in polar way (spherical oordinates) and get
$$
\eqalign{
  & \left| {\,{\bf a} \wedge {\bf b}\,} \right|
 = \left| {\,{\bf a}\,} \right|\left| {\,{\bf b}\,} \right|\sin \left( {\angle {\bf a},{\bf b}} \right)
 = a\,b\sin \left( {\angle {\bf a},{\bf b}} \right) =   \cr 
  &  = \left| {\,{\bf a}\,} \right|\left| {\,{\bf b}_{\, \bot {\bf a}} \,} \right| =   \cr 
  &  = a\sqrt {b^{\,2}  - \left( {{\bf b} \cdot {{\bf a} \over a}} \right)^{\,2} }
  = \sqrt {a^{\,2} b^{\,2}  - \left( {{\bf b} \cdot {\bf a}} \right)^{\,2} }  \cr} 
$$
where
$$
\left\{ \matrix{
  {\bf b} = {\bf b}_{\,\backslash \backslash {\bf a}}  + {\bf b}_{\, \bot {\bf a}}  \hfill \cr 
  {\bf b}_{\,\backslash \backslash {\bf a}}  = \left( {{\bf b} \cdot {\bf a}/a} \right){\bf a}/a \hfill \cr 
  {\bf b}_{\, \bot {\bf a}}  = {\bf b} - {\bf b}_{\,\backslash \backslash {\bf a}}
  = {\bf b} - \left( {{\bf b} \cdot {\bf a}/a} \right){\bf a}/a \hfill \cr 
  \left| {{\bf b}_{\,\backslash \backslash {\bf a}} } \right|^{\,2}
  = \left( {{\bf b} \cdot {\bf a}/a} \right)^{\,2}
  = b^{\,2} \cos ^{\,2} \left( {\angle {\bf a},{\bf b}} \right) \hfill \cr 
  \left| {{\bf b}_{\, \bot {\bf a}} } \right|^{\,2}
  = \left( {{\bf b} - \left( {{\bf b} \cdot {\bf a}/a} \right){\bf a}/a} \right) \cdot
 \left( {{\bf b} - \left( {{\bf b} \cdot {\bf a}/a} \right){\bf a}/a} \right) =  \hfill \cr 
   = \left| {\bf b} \right|^{\,2}  - 2\left( {{\bf b} \cdot {\bf a}/a} \right)^{\,2}
  + \left( {{\bf b} \cdot {\bf a}/a} \right)^{\,2}
  = b^{\,2}  - \left( {{\bf b} \cdot {\bf a}/a} \right)^{\,2}  =  \hfill \cr 
   = b^{\,2} \sin ^{\,2} \left( {\angle {\bf a},{\bf b}} \right) \hfill \cr}  \right.
$$
So in your case,
$$
\eqalign{
  & {\bf a} = \left( {2, - 8, - 1, - 1} \right)\quad {\bf b} = \left( { - 4, - 18, - 1, - 3} \right)  \cr 
  & {\bf w} = {\bf a} \wedge {\bf b}
 =  - 68{\bf e}_{\,1} {\bf e}_{\,2}  - 6{\bf e}_{\,1} {\bf e}_{\,3}  - 10{\bf e}_{\,1} {\bf e}_{\,4}
  - 10{\bf e}_{\,2} {\bf e}_{\,3}  + 6{\bf e}_{\,2} {\bf e}_{\,4}  + 2{\bf e}_{\,3} {\bf e}_{\,4}  \cr} 
$$
and the area of the parallelogram defined by $A,B,C$ will be given by
$$
\left| {\bf w} \right| = A_{\,ABCA'}
  = \sqrt {68^{\,2}  + 6^{\,2}  + 10^{\,2}  + 10^{\,2}  + 6^{\,2}  + 4^{\,2} }  = 70
$$
while that of the triangle will be half of that.
