How do I calculate the *signed* area of a triangle in 3D space? The signed area of a triangle is often given simply as
\begin{equation}
  A\left( \Delta \right) = \frac{1}{2}\left|\begin{matrix}
    x_{0} & y_{0} & 1  \\
    x_{1} & y_{1} & 1  \\
    x_{2} & y_{2} & 1
\end{matrix}\right|
\end{equation}
for $V_i = \left(x_i,y_i\right)$.
This, of course, only works for vertices located on a 2D plane.
What is the analog for the signed area of a triangle in 3D space, having $x,y,z$ coordinates?
 A: In 2D the sign of the determinant defining the area of the triangle
is determined by the order in which you consider the vertices, i.e.
the three rows in your determinant
$$
\left| {\matrix{
   {x_0 } & {y_0 } & 1  \cr 
   {x_1 } & {y_1 } & 1  \cr 
   {x_2 } & {y_2 } & 1  \cr 
 } } \right|
$$
or otherwise the two vertical vectors corresponding to sides
departing from $(x_0,y_0)$ which appear in the related determinant
$$
\left| {\matrix{
   {x_1  - x_0 } & {x_2  - x_0 }  \cr 
   {y_1  - y_0 } & {y_2  - y_0 }  \cr 
 } } \right|
$$
If you swap two of them you change the sign.
The sign defines the chirality
according for instance to the right-hand rule well known in physics.
Now if you  translate or rotate the triangle as a "rigid body", by multiplying by a rotation matrix with
determinant $+1$, the chirality is not changed.
So, given a triangle in 3D, once you establish the order of vetices/the sides, it will have a precise signed area, which is the same
that you have once you rotate it onto the $x-y$ plane.
