Equation of a vertical plane given $2$ points A vertical plane passes through points $(1, -1, 1)$ and $(2, 1, 1)$. 
With three points on a normal plane, I just found two vectors and found the normal by cross-product of the vectors, but I'm not sure what to do here. 
Help, please.
 A: There are plenty of ways you can represent a plane mathematically. One of the most succinct ones is using the normal and a position vector as
$$\hat{n}\cdot (\vec{x}-\vec{c})=0$$
Where $\hat{n}$ is the vector normal to the plane and $\vec{c}$ is any vector in the plane.
Now in your case you have two points and you know that the plane is vertical, which I'm assuming it means that is perpendicular to the $x-y$ plane. With that information you can find a third point contained in the plane for example by setting the $z$ coordinate of any of the given vectors to 0.
Now with 3 vectors $v_1=(1,-1,1);\; v_2=(2,1,1);\; v_3=(1,-1,0)$ you can calculate the normal vector using the cross product of any two vector lying in the plane, for example $v_1-v_2$ and $v_2-v_3$.
This gives you that
$$\hat{n}=(v_1-v_2)\times (v_2-v_3) = (-2,1,0)$$
Which makes sense as you know that the $z$ coordinate must be 0 in order to be a vertical plane.
With that information you can now write your plane equation as
$$(-2,1,0)\cdot (\vec{x}-(1,-1,1))=0$$
Or if you expand it with $\vec{x}=(x,y,z)$ you can get
$$2x-y-3=0$$
A: Hint: Since the plane is vertical, it is parallel to the vector $(0, 0, 1)$.
