line parallel to plane, but not on plane. I need to find a plane that goes through the points $A=(2,0,2)$ and $B=(4,1,0)$, that is parallel to the line?
$$r(t) = (0,3,-2) + t\langle1,-1,1\rangle$$
or if you want it in parametric equations:
$$x = t, \ y = 3 - t, \ z = t - 2.$$
How do I find a plane that goes through two points? and how do I decide if it is parallel to the line?
 A: Notice that if the plane is parallel to L, then the vector normal to the plane is then perpendicular to the line. Find the/ (a) vector normal to the plane, then you have two points in the plane, and you're done.
And there are infinitely-many planes that go through any two given points; there are infinitely-many planes that even go through a given line. Once you're given a vector normal to the plane, and two points in the plane, you're done ( although, given two points, you can find N using their cross-product. )
If the line lies in the plane, you can translate the plane to avoid containing the line.
A: The plane is parallel to both the line $AB$ and to the given line, so it is normal to the cross product of vectors along these lines. Knowing the normal and a point (let $A$), finding the equation of the plane is straightforward.
A: In addition to the answers already given, you can consider that
if the line is parallel to the plane, you can "transport" the vector defining the line onto the plane, in particular you can "fix" it on $B$ (or on $A$) and determine a third point $C=(4,1,0)+(1,-1,1)=(5,0,1)$
