$$
x = -2-4t,~y = 3-2t,~z = 1+2t\\ 2x+y-z=5
$$
I know that in order to be perpendicular the vectors should be orthogonal so their dot product should equal zero.
The normal vector is $(2,1-1)$, and the vector for the lines is $(-4,-2,2)$, so
$$
(2,1,-1) \cdot (-4,-2,2) = -8-2-2 = -12
$$
Since the dot product is not equal to $0$ the line isn't perpendicular to the plane. Am I on the right track?
Best Answer
A normal vector to the plane is $ \ \langle \ 2, 1, -1 \ \rangle \ $ , which is perpendicular to the plane. Any scalar multiple of this vector is, as well. The direction vector of the line is $ \ \langle \ -4, -2, 2 \ \rangle \ $ . These you have found correctly.
Since $ \ \langle \ -4, -2, 2 \ \rangle \ = \ -2 \ \langle \ 2, 1, -1 \ \rangle \ $ , these two vectors are "parallel", that is to say, their directions lie along the same line in three-dimensional space. (Some people would call them "anti-parallel".) So both vectors are perpendicular to the given plane.