The area of a triangle in $\mathbb{R}^3$ is $\sqrt{S_1^2+S_2^2+S_3^2}$, where each $S_i$ is the area of a projection onto a coordinate plane

Question

Prove that if the three vertices of triangle $ABC$ are $A(x_1,y_1,z_1),B(x_2,y_2,z_2),C(x_3,y_3,z_3)$,
then the area of triangle $ABC=S=\sqrt{S_1^2+S_2^2+S_3^2}$, where $$S_1=\frac{1}{2}\begin{vmatrix}y_1&z_1&1\\y_2&z_2&1\\y_3&z_3&1\end{vmatrix} \qquad S_2=\frac{1}{2}\begin{vmatrix}x_1&z_1&1\\x_2&z_2&1\\x_3&z_3&1\end{vmatrix} \qquad S_3=\frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}$$

Here, $S_1$, $S_2$, $S_3$ are the areas of the projections of $\triangle ABC$ onto the $yz$, $zx$, $xy$ planes, respectively.

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This formula is given in my book, but I do not know how this formula is derived. I tried, but failed. I googled and found this, but could not understand it: https://www.johndcook.com/blog/2016/08/23/area-of-a-triangle-and-its-projections/

Please help.