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i've checked multiple places for a general formula to follow including here:

Find an equation of the plane passing through 2 points and perpendicular to another plane

I know it asks the same question but it does not explain how it got to the second indented box, I can't seem to figure out how to derive the info in it and it seems to unique to answer the question in general, so what is the general step by step formula for answering this question? my sample question I'm trying to answer is right here:

Find an equation for the plane which is perpendicular to the plane x-2y+2z=3 and passing through the points (2, -1, 1) and ( 1,3 ,0)

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J L
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Essentially you need the normal to the plane, this can be found if you have two vectors parallel to the plane. One is given by the difference of the points, $(2,-1,-2)-(1,3,0)$ the second is the normal to $x-2y+2z=3$ which is $(1,-2,2)$. Now you just take the cross product of these vectors to get the normal. To finish you plug in one point.

Rene Schipperus
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  • ok im with you up to the point where you take cross product to get normal right before the finish. but plug one point into what? how does that give an equation? – J L Jun 30 '14 at 04:42
  • Lets say the normal is $(1,2,3)$ then the equation will be $x+2y+3z=a$ now we need to find $a$, we can use a point if we have it. – Rene Schipperus Jun 30 '14 at 04:48
  • by a point that "we have" do you mean one of the points the equation passes through like (2, -1, 1) and ( 1,3 ,0)? and then how will the final equation look? – J L Jun 30 '14 at 04:56
  • Yes that is it. – Rene Schipperus Jun 30 '14 at 04:59
  • I'm also asking what the general form of the equation will look like based on the inputed point – J L Jun 30 '14 at 05:05