# Finding the vector equation of a plane in 3 space

I'm working on a question that asks me to find the vector equation of the plane in $$\mathbb R^3$$ that contains the point P(-1,6,0) and is parallel to the plane x+2y-z=5. I'm trying to make sure my answer is right versus the answer that I was provided with, which I think is wrong.

I pulled out the normal vector n = (1,2,-1). I proceeded to find two vectors v and u that are orthogonal to n but are not parallel to each other. I followed the theorem that states that when the dot product of 2 vectors equals zero, those vectors are orthogonal. I came up with v = (1,1,3) and u = (4,1,6). So I ended with the vector equation (x,y,z) = (-1,6,0) + t(1,1,3) + s(4,1,6).

It's my understanding that there are a number of different answers for this problem as there are a number of different vectors, besides the ones I found, that are orthogonal to the normal vector. The answer I was provided for this question stated that the vector equation is (x,y,z) = (-1,6,0) + t(3,-4,1) + s(1,-4,-1). I'm fairly certain that this answer is incorrect as the vectors in the vector equation are not orthogonal to the normal equation. Is my thinking correct?

The plane parallel to $$x+2y-z=5$$ containing $$(-1,6,0)$$ is $$x+2y-z=-1+2(6)-0=11$$.
Your parametrization satisfies that: $$(-1+t+4s)+2(6+t+s)-(3t+6s)=11$$.
The other answer does not: $$(-1+3t+s)+2(6-4t-4s)-(t-s)\not=11$$ generally.