TITLE: Find parametar λ if plane and line don't intersect
QUESTION [5 upvotes]: How can I find λ, if I have line $3\lambda x-2=3y+1=\lambda z$ and plane $\lambda x-3y+2z-3=0$ and they don't intersect. The given solution says that λ=3, but I don't have any idea how can I come to that solution. Small hint would be helpful...

REPLY [4 votes]: $$3\lambda x-2=3y+1=\lambda z→\frac{x-2/3\lambda}{1}=\frac{y-1/3}{\lambda}=\frac{z}{3}$$
So $(1,\lambda,3)$ is the vector direction. On the other hand $(\lambda,-3,2)$ is the normal vector to the plane. 
If there is no intersection it means that the line is parallel to the plane and then those vectors are orthogonal.
$$(1,\lambda,3)\cdot (\lambda,-3,2)=\lambda-3\lambda+6=0→\lambda=3$$