Find distance from line k given parametric equation of: $x=2+t, y=-3+2t, z=2-t, t\in\mathbb{R}$ from plane $\pi:2x+y+4z=0$. My solution:
$$2(2+t)+(-3+2t)+4(2-t)$$
$$4+2t-3+2t+8-4t$$
$$9=0$$
Contradiction, so no solutions, line and plane are parallel.
Its first time where I have such an example where equation is contradiction, I followed some example from internet and those are my next steps:
$$d(l,\pi)=\frac{|9|}{\sqrt{1+2^2+1}}=\frac{9\sqrt{6}}{6}$$
I am not sure about this method, is this distance value right?
Maybe there's better way to calculate it?
 A: The direction vector $v$ of the line and the plane normal $n$ are orthogonal:
$$\begin{pmatrix}1\\2\\-1 \end{pmatrix} \cdot \begin{pmatrix}2\\1\\4 \end{pmatrix} = 0$$
and thus the line is parallel to the plane and all points on the line have the same distance to the plane.
The formula for the distance from a point $p$ to a plane $\pi:n\cdot x + d = 0$ is 
$$\frac{|p \cdot n + d|}{||n||}$$
For your plane:
$$\frac{\left|\begin{pmatrix}2\\-3\\2 \end{pmatrix} \cdot \begin{pmatrix}2\\1\\4 \end{pmatrix} + 0\right|}{\sqrt{2^2+1^2+4^2}} = \frac{9}{\sqrt{21}}$$
So you accidentally normalized the the wrong vector (line direction vector instead of plane normal). Besides that your calculations are correct.
A: Take any point on line and find length of perpendicular dropped from that point onto the plane. Let point is (2,-3,2),(t=0). So the formula for perpendicular distance is $\frac {|ax_0+by_0+cz_0+d_0|} {\sqrt{a^2+b^2+c^2}}$. Using this the distance is $\frac {|2×2+1×(-3)+4×2|} {\sqrt{2^2+1^2+4^2}}$
