Find parametar λ if plane and line don't intersect 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...
 A: Hint:
The line is the intersection of the two planes
$$
3 \lambda x - 2 = 3 y + 1 \iff 3 \lambda x - 3 y = 3 \\
3y + 1 = \lambda z \iff 3 y - \lambda z = -1 
$$
Then we take the plane
$$
\lambda x - 3 y + 2z - 3 = 0
$$
into the consideration of common intersections. 
This is equivalent to analyzing the solutions of the inhomogeneous linear system $A u = b$ with the augmented matrix
$$
[A | b] =
\left[
\begin{array}{rrr|r}
3 \lambda & -3 & 0 & 3 \\
0 & 3 & -\lambda & - 1 \\
\lambda & -3 & 2 & 3
\end{array}
\right]
$$
Solution:
This can be transformed into
$$
\left[
\begin{array}{rrr|r}
0 & 6 & -6 & -6 \\
0 & 3 & -\lambda & - 1 \\
\lambda & -3 & 2 & 3
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
0 & 1 & -1 & -1 \\
0 & 3 & -\lambda & -1 \\
\lambda & -3 & 2 & 3
\end{array}
\right]
\to \\
\left[
\begin{array}{rrr|r}
0 & 1 & -1 & -1 \\
0 & 0 & 3 -\lambda & 2 \\
\lambda & -3 & 2 & 3
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
\lambda & -3 & 2 & 3 \\
0 & 1 & -1 & -1 \\
0 & 0 & 3 -\lambda & 2 
\end{array}
\right]
$$
We see that the choice $3 - \lambda = 0$ leads to an inconsistency in the last equation ($0 \cdot x + 0 \cdot y + 0 \cdot z = 2$), and thus no solution of the system.
A: $$3\lambda x-2=3y+1=\lambda z→\frac{x-2/3\lambda}{1}=\frac{y-1/3}{\lambda}=\frac{z}{3}$$
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.
$$(1,\lambda,3)\cdot (\lambda,-3,2)=\lambda-3\lambda+6=0→\lambda=3$$
A: You could try looking for the point where the line intersects the plane,
supposing that such a point did exist.
In other words, see what would happen if $\lambda$ were some known
value that is not a solution of the problem.
From the specification of the line, you have equations that let you solve for
any variable $x,y,z$ in terms of either of the others.
Choose one variable, solve for the other two, and plug those solutions
into the equation of the plane.
Now you have an equation that is true for a point on the line
and a point on the plane, so it's true at the intersection point.
The equation you have just found is a linear equation in one variable,
so you can solve for that one variable.
This will always work except in one case, and that case occurs
when you don't actually even have one variable left in the equation,
that is, plugging in the solutions of the other two variables
gets you something equivalent to $1 = 0.$
That one case occurs for a particular value of $\lambda,$
and that's the value you want.
