Equation of a plane involving a known line and a known point

[Oxygenate]

Homework Statement

Find the equation of the plane which contains the line r(t) = (2-t, 1+t, t) and the point (1,0,1).

The Attempt at a Solution

Since r(t) = (2-t, 1+t, t), we know that vector a, which is parallel to the line, is (-1, 1, 1).
I'm assuming that since the point is (1,0,1), that means if we look to that point from the origin (0,0,0) we will have vector b = (1,0,1).
Okay so with two vectors a and b, I can find a vector n such that n is orthogonal to vectors a and b. So a x b = (1-0,1+1,0-1) = (1,2-1) = vector n.
The scalar equation of a plane is a(x-x₀) + b(y-y₀ + c(z-z₀). So x + 2y -z = 0...

But somehow I have a feeling that my result is wrong. Is there supposed to be a "y" at all in the equation for this plane? Please advise. Thanks in advance!

 

[Mark44]

The point B(1, 0, 1) is in the plane, but the vector OB isn't. From your equation for the line you can easily find two points in the plane, say (2, 1, 0) and (1, 2, 1), just by supplying values of t.

Form vectors between one of the points and each of the other two. Now cross these vectors to get a vector normal to both of these. Use this normal vector and anyone of the points to get the equation of the plane.

You didn't show your work in getting the cross product a X b, so the normal you show is somewhat suspect.