I have a point $(9,5,0)$ and a triangle with points $(1,1,0), (3,3,1), (6,1,0)$, let's label them as $A,B,C$ respectively. In order to get the normal vector, I do the cross product of two vectors. If I do $AB \times AC$ I get $n = \left\langle 0,-5,10\right\rangle$. With this I determine the equation of the plane is $5y - 10z - 5 = 0$. If I project my point onto the plane with this normal vector in mind, the point on the plane I find is $(9, 21/5, 8/5)$. HOWEVER, if I initially choose to cross two different vectors, I get something different.
Example: If I instead choose to do $BA \times BC$ I get $n = \left\langle 0,5,-10\right\rangle$. Now this in a way makes sense to me since it is the inverse of the other normal I found. Basically the other half of the line through the plane, right? If I continue, I find the equation of the plane to be $-5y + 10z + 5 = 0$. This leads me to a projected point of $(9, -29/5, -8/5)$. Why do I get two different points based on which normal vector I choose to solve with? I know the correct answer to this problem, one is right and one is wrong. I am trying to program a general solution to this type of problem. How do I choose the correct normal vector?
Please let me know if I am doing something wrong.