I understand how one can calculate a plane equation (ax+by+cz=d) from three points but how can you go in reverse?
How can you calculate arbitrary points from a plane equation?
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$\begingroup$From your comment I finally understood what you are looking for:
If you have a plane defined by $a x + b y + c z = d$ then you also have the following properties:
Plane normal direction: $$\hat{n} = \begin{pmatrix} \frac{a}{\sqrt{a^2+b^2+c^2}} \\ \frac{b}{\sqrt{a^2+b^2+c^2}} \\ \frac{c}{\sqrt{a^2+b^2+c^2}} \end{pmatrix}$$
Point on plane closest to the origin (position of plane)$$ \vec{r} = \begin{pmatrix} \frac{a d}{a^2+b^2+c^2} \\ \frac{b d}{a^2+b^2+c^2} \\ \frac{c d}{a^2+b^2+c^2} \end{pmatrix} $$
Distance of plane from the origin$$ r = \frac{d}{\sqrt{a^2+b^2+c^2}} $$
Directions along the plane (not unit vectors), and perpendicular to $\hat{n}$.$$ \begin{align} \hat{e}_1 & = \begin{pmatrix} c-b \\ a-c \\ b-a \end{pmatrix} & \hat{e}_2 & = \begin{pmatrix} a (b+c)-b^2-c^2 \\ b (a+c) -a^2-c^2 \\ c (a+b)-a^2 - b^2 \end{pmatrix} \end{align} $$
You can verify that $\hat{e}_1 \cdot \hat{n} =0$, $\hat{e}_2 \cdot \hat{n}=0$ and $\hat{e}_1 \cdot \hat{e}_2 =0$, where $\cdot$ is the dot (inner) product.
Confirmation via GeoGebra
NOTES: Please edit the equation to make it clear you are looking for the plane properties when given a plane in equation form.
$\endgroup$ 5 $\begingroup$You pick arbitrarely two values $a,b$ for $x$ and $y$, you plug them in the equation, you solve it for $z$ and obtain a value $c$. The point $(a,b,c)$ is a point of your plane.
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