Geometric intuition for the complex shoelace formula The complex shoelace formula for the signed area of a triangle with vertices given by the complex numbers $a, b, c$ is $$\frac{i}{4}
    \begin{vmatrix}
    1 & 1 & 1 \\
    a & b & c \\
    \overline{a} & \overline{b} & \overline{c} \\
    \end{vmatrix}
$$
I have seen the algebraic proof for this formula using elementary row operations and the multilinear nature of the determinant, however that style of proof appears to imply that the simplicity of the final result (i.e. a determinant involving only conjugates) is a coincidence; furthermore it provides little intuition.
I am looking for a way to understand this formula through a geometric/intuitive argument (not necessarily rigorous) in order to gain a deeper understanding rather than just accepting that the algebra works out.

Consider the case $\,a=0\,$, first, where the proposed formula reduces to:
$$
S_{OBC} = \frac{i}{4}
\begin{vmatrix}
 1 & 1 & 1 \\
 0 & b & c \\
 \overline{0} & \overline{b} & \overline{c} \\
\end{vmatrix}
=
\frac{i}{4}
\begin{vmatrix}
 b & c \\
 \overline{b} & \overline{c} \\
\end{vmatrix}
= \frac{i}{4}\left(b \bar c - \bar b c\right)
= -\,\frac{1}{2}\,\operatorname{Im}(b \bar c) \tag{1}
$$
Let $\,b = |b| e^{i \beta}\,$, $\,c = |c| e^{i\gamma}\,$, then $\,b \bar c = |b|\cdot|c|\cdot e^{i(\beta-\gamma)}=\color{blue}{|b|}\cdot\color{blue}{|c|}\cdot \left(\cos(\beta-\gamma)+ i \color{red}{\sin(\beta-\gamma)}\right)\,$, so $\,(1)\,$ is equivalent to the familiar triangle (signed) area formula $\,S_{OBC} = \frac{1}{2}\,\color{blue}{OB} \cdot \color{blue}{OC} \cdot \color{red}{\sin \widehat{BOC}}\,$.
But area is, of course, invariant under translations, so it follows that in the general case:
$$
S_{ABC} = \frac{1}{2}\,AB \cdot AC \cdot \sin \widehat{BAC} = -\,\frac{1}{2}\,\operatorname{Im}\left((b-a) \overline{(c-a)}\right) \tag{2}
$$
The latter is equivalent to the posted formula via elementary column operations:
$$
\begin{vmatrix}
 1 & 1 & 1 \\
 a & b & c \\
 \overline{a} & \overline{b} & \overline{c} \\
 \end{vmatrix}
\;=\;
\begin{vmatrix}
 1 & 0 & 0 \\
 a & b-a & c-a \\
 \overline{a} & \overline{b-a} & \overline{c-a} \\
 \end{vmatrix}
$$
