A simple question on position vectors. Is there any unit vector normal to all the following $3$ position vectors? $$a=\langle 2,-3,1\rangle,\quad b=\langle3,2,-4\rangle,\quad c=\langle-1,2,-2\rangle.$$ I know the answer to this. But the answer I have is apparently not the right one. I spent a whole hour arguing with my teacher on this trivial question as he seemed to have a different understanding of position vectors.  What I did: $$a\times b=\langle10,11,13\rangle=p \\ p\cdot c\neq0.$$ Therefore, position vectors $a,b,c$ do not lie on the same plane and do not have a unit normal vector common to all of them. What he says we should do: Shift the origin to '$a$' and take the position vectors $ab$ and $ac$. Then take a cross product and divide it by its magnitude to get the unit vector perpendicular to position vectors $a,b$ and $c$. I told him the vectors $ab$ and $ac$ weren't really the original position vectors and so that was wrong. But he kept stressing the point that they are position vectors and hence that can be done. He didn't seem to get my question. So who is right? I searched a lot online and nowhere did they seem to use his way. This is my last resort as I didn't really want to put up such a question here unless there was no other way.
You are absolutely right. If he does not agree ask him to take dot product of what he found out with $\vec a,\vec b ,\vec c$.

And if he still does not agree, ignore him.
your teacher is right: any three points lie on a plane. Your solution assumes a plane which goes through the origin.

What your teacher does is intuitively clear: when three points lie on a plane, it does not matter whether you move all of them by the same amount. They should still be on a plane (albeit a different one). Thus when your teacher subtracts a from all points, he ends up with (0,0), ab and ac. Next, he constructs a plane through the origin so (0,0) is in there. the normal of the plane is defined by ab x ac. and thus ab and ac are on the plane.