Why isn't the orthogonal vector to a direction vector of a plane not necessarily perpendicular to such plane? I had made a question, and the problem with my exercise was that I was trying to calculate a vector perpendicular to some plane in $\mathbb{R}^3$: given one line $L$ inside the plane, I grabbed the direction vector $(-2,0,3)$ of this line and then solve
$$(x,y,z) \cdot (-2,0,3) = 0$$
To get an orthogonal vector to the plane. But I was told that

Not all vector perpendicular to $(−2,0,3)$ gives you the normal vector
  of the plane.

However I don't really see why. If $(-2,0,3)$ is a direction vector of the plane, why wouldn't a vector perpendicular to $(-2,0,3)$ necessarily be also perpendicular to said plane? In $\mathbb{R}^3$, at least, I can't really visualize why not.

The vector perpendicular to your line will be just that, perpendicular to your line; nothing more is guaranteed. To be perpendicular to the plane, you must be perpendicular to all vectors (or lines) in the plane at once; there is only one direction (and its opposite) which does that. The vector you found perpendicular to $(2,0,-3)$ could very well be inside the plane; there is certainly a direction within the plane that is perpendicular to your line. Or your perpendicular vector could happen to be a normal vector to the plane. Or anything in between.
One really cannot tell more just knowing that the vector is perpendicular to $(2,0,-3)$.
