Prove that the line x = 1 + 2t, y = -1 + 3t, z = 2 + 4t is parallel to the plane x - 2y + z = 6.
What is the distance between the line and the plane?
not that the vector in the direction of the line is <2,3,4>. we want this to be perpendicular to the normal to the plane, which is <1,-2,1>. as you can see, that does not happen
if you want to find the distance between a plane and a line parallel to it, use the formula:
$\displaystyle D = \frac {|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$
this gives the distance between a point $\displaystyle (x_0, y_0, z_0)$ and the plane $\displaystyle ax + by + cz + d = 0$, just make sure $\displaystyle (x_0, y_0, z_0)$ is a point on the line