Determine whether the line and plane are perpendicular $$
x = -2-4t,~y = 3-2t,~z = 1+2t\\ 2x+y-z=5
$$
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?
 A: 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.
A: the dot product is 0 means that the normal vector and the line are not orthogonal. but still, there is a chance that the line is perpendicular to the plane and in order to do validate that, you have to calculate the cross product between the normal vector and the parallel vector from line, and accordingly if the value is 0, then the normal vector and the line are parallel and thus the line will be perpendicular to the plane, otherwise, it is neither parallel (proved by dot product), and neither orthogonal (proved by cross product)
