Problem source

State carefully the conditions which the fixed vectors $\mathbf{a,b,u}$ and $\mathbf{v}$ must satisfy in order to ensure that the line $\mathbf{r=a+}\lambda\mathbf{u}$ intersects the line $\mathbf{r=b+\mu}\mathbf{v}$ in exactly one point. 
Find the two values of the fixed scalar $b$ for which the planes with equations 
\[
\left.\begin{array}{c}
x+y+bz=b+2\\
bx+by+z=2b+1
\end{array}\right\} \tag{*}
\]
do not intersect in a line. For other values of $b$, express the line of intersection of the two planes in the form $\mathbf{r=a}+\lambda\mathbf{u},$ where $\mathbf{a\cdot u}=0$. 
Find the conditions which $b$ and the fixed scalars $c$ and $d$ must satisfy to ensure that there is exactly one point on the line
\[
\mathbf{r=}\left(\begin{array}{c}
0\\
0\\
c
\end{array}\right)+\mu\left(\begin{array}{c}
1\\
d\\
0
\end{array}\right)
\]
whose coordinates satisfy both equations $(*)$.

Solution source

There are two requirements (assuming they are lines not fixed points):

1. They cannot be parallel, ie $\mathbf{u} \neq \lambda \mathbf{v}$ for any $\lambda$
2. They must lie in the same plane, ie $(\mathbf{b}-\mathbf{a})\cdot (\mathbf{u} \times \mathbf{v}) = 0$

The planes will not intersect in a line if they are either parallel and separate or parallel and the same.

If $b = 1$ or $b=-1$ the planes are parallel.

\begin{align*}
&& (x+y) + b z &= b+ 2\\
&&b(x+y) + z &= 2b + 1 \\
\Rightarrow && (1-b^2)z &= 2b+1 - b^2 -2b \\
&&&= 1-b^2 \\
\Rightarrow && z &= 1 \\
&& x+ y &= 2 \\
\end{align*}

Therefore our line is $\mathbf{r} = \begin{pmatrix} 1+t \\ 1-t \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + t  \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} $

We must have:

$d \neq -1$ to ensure that the lines aren't parallel.

We must also have:

\begin{align*}
0 &= \left ( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} -\begin{pmatrix} 0 \\ 0 \\ c \end{pmatrix}\right) \cdot \left (  \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \times  \begin{pmatrix} 1 \\ d \\ 0 \end{pmatrix} \right) \\
&=  \begin{pmatrix} 1 \\ 1 \\ 1-c \end{pmatrix} \cdot  \begin{pmatrix} 0 \\ 0 \\ d+1 \end{pmatrix} \\
&= (1-c)(d+1)
\end{align*}
So $c =1$