<s>[INST] 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...
 [/INST] 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.
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