keywords: Math, Plane equation given 3 points

Here's one way to get the requisite plane:

1. Get two different vectors which are in the plane, such as $$B−A=(3,0,−3)$$ and $$C−A=(3,3,3)$$.
2. Compute the cross product of the two obtained vectors: $$(B−A)×(C−A)=(9,−18,9)$$. This is the normal vector of the plane, so we can divide it by 9 and get $$(1,−2,1)$$.
3. The equation of the plane is thus $$x−2y+z+k=0$$. To get k, substitute any point and solve; we get $$k=−6$$.

The final equation of the plane is $$x−2y+z−6=0$$.

Origin:
Equation of a plane passing through 3 points
https://math.stackexchange.com/a/2686620/601445

##### Reference

Equation of a plane passing through 3 points
https://math.stackexchange.com/questions/2686606/equation-of-a-plane-passing-through-3-points

General equation of a plane through 3 points.
https://math.stackexchange.com/questions/1449049/general-equation-of-a-plane-through-3-points

Equation of a Plane Passing Through 3 Three Points
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