\documentclass[fleqn]{article}

\usepackage{amsmath}

\usepackage{marvosym}

\usepackage{tikz}

\usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools

\begin{document}

You wish to have a coordinate system that 

\begin{enumerate}

 \item preserves shapes and\label{preserve}

 \item has $\vec e_y$ point east, $\vec e_z$ point north and $\vec e_x$ point

 south west.\label{directions}

\end{enumerate}

The first requirement means that the coordinate axes are orthogonal,

\begin{equation}\label{eq:orthogonality}

 \vec e_x\cdot \vec e_y=\vec e_x\cdot \vec e_z=\vec e_y\cdot \vec e_z

=0\;.

\end{equation}

So we wish to find a two--dimensional projection of these vectors that fulfill

the requirement \ref{directions}. Decompose the vectors in two--dimensional

projections on the paper plane $\vec e_i^{(\|)}$ and the orthogonal complements

$\vec e_i^{(\perp)}$. Clearly, the $\vec e_i^{(\perp)}$ are just

one--dimensional objects, which we will just call $e_i^{(\perp)}$. Requirement \ref{directions} implies that

\begin{equation}

 \vec e_x^{(\|)} \cdot \vec e_y^{(\|)}=

\vec e_x^{(\|)} \cdot \vec e_z^{(\|)}=:\xi\ne0\;.

\end{equation}

Due to the orthogonality relations \eqref{eq:orthogonality}, this means that

\begin{equation}

 e_x^{(\perp)} \cdot e_y^{(\perp)}=

 e_x^{(\perp)} \cdot e_z^{(\perp)}=-\xi\ne0\;.

\end{equation}

None of the $e_i^{(\perp)}$ may vanish as otherwise there won't be an $x$--axis,

and

\begin{equation}

e_y^{(\perp)}=e_z^{(\perp)}=-\frac{\xi}{e_x^{(\perp)}}\;. 

\end{equation}

However, requirement \ref{directions} 

implies that $\vec e_y^{(\|)} \cdot \vec e_z^{(\|)}=0$, so

\begin{equation}

 \vec e_y\cdot \vec e_z=\vec e_y^{(\|)} \cdot \vec e_z^{(\|)}

 +e_y^{(\perp)} \cdot e_z^{(\perp)}=0+\left(\frac{\xi}{e_x^{(\perp)}}\right)^2

 \ne0\;.\qquad \text{\Huge\Lightning}

\end{equation}

%\clearpage

Having seen that the desired projection is not orthonormal, one may ask whether

it is possible to use \texttt{3dtools} for such projections. The answer is that

this is in principle possible. You can define your own \texttt{screendepth}

function, in this very case this is not even necessary. However, in general it

won't be easy to guess an appropriate fuction.

​

\begin{figure}[htb]

 \centering

 \begin{tikzpicture}

  \pgfmathdeclarefunction*{screendepth}{3}{\pgfmathparse{0.2*#1+0.2*#2+#3}}

  \path foreach \Xa/\Ya in {-1/A,1/B}

   {foreach \Xb/\Yb in {-1/A,1/B}

   {foreach \Xc/\Yc in {-1/A,1/B}

   {(\Xa,\Xb,\Xc) coordinate (\Ya\Yb\Yc)}}} ;

  \tikzset{3d/polyhedron/.cd,

         fore/.append style={fill opacity=0.5},

         edges have complete dashes,

         draw face with corners={{(AAA)},{(AAB)},{(ABB)},{(ABA)}},

         draw face with corners={{(AAA)},{(AAB)},{(BAB)},{(BAA)}},

         draw face with corners={{(AAA)},{(BAA)},{(BBA)},{(ABA)}},

         draw face with corners={{(BAA)},{(BAB)},{(BBB)},{(BBA)}},

         draw face with corners={{(ABA)},{(ABB)},{(BBB)},{(BBA)}},

         draw face with corners={{(AAB)},{(BAB)},{(BBB)},{(ABB)}}}

 \end{tikzpicture} 

\end{figure}

\end{document}