The plane \(\pi\) has equation \(\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}+ \mu \mathbf{c}\) and the plane \(\pi'\) has equation \(\mathbf{r} = \mathbf{d}+ \lambda'\mathbf{e}+\mu'\mathbf{f}\), where \(\lambda\), \(\mu\), \(\lambda'\) and \(\mu'\) are scalar parameters. It is given that

\[\mathbf{a}=\begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix}, \mathbf{b}=\begin{pmatrix}2\\ -1 \\ -1\end{pmatrix}, \mathbf{c} = \begin{pmatrix}-1 \\ 2 \\ 1\end{pmatrix}, \mathbf{d} = \begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}, \mathbf{e} = \begin{pmatrix}p \\ q \\ r\end{pmatrix}, \mathbf{f} = \begin{pmatrix}q\\ r\\ p\end{pmatrix},\]

\(\big \vert\mathbf{e}\big \vert=\big \vert\mathbf{f}\big \vert = \sqrt{30}\) and that \(p\) is negative. The planes \(\pi\) and \(\pi'\) are parallel. Find \(p\), \(q\) and \(r\).

If \(\pi\) and \(\pi'\) are parallel, what must be true about the normal of either plane?

How could we use the vector and scalar products here?

Show that \(\pi'\) is the reflection of \(\pi\) in the origin.

If \(P\) is the point \((x,y,z)\), what are the coordinates of the reflection of \(P\) in the origin?

The line \(L\) has equation \[\dfrac{x+1}{9}=\dfrac{y-1}{-6}=\dfrac{z}{-5}\] and lies in \(\pi\). The point \(P\) is in \(\pi\) with parameters \(\lambda = l\) and \(\mu = m\), and \(P'\) is in \(\pi'\) with \(\lambda' = l\) and \(\mu' = m\). The point \(Q'\) is the reflection of \(P\) in the origin. Show that when \(P\) lies on \(L\) the direction of \(P'Q'\) is independent of the position of \(P\) on \(L\).

If \(P\) is on \(L\), can we find the coordinates of \(P\) in terms of a parameter \(\alpha\)?

Can we now find the coordinates of \(P'\) and \(Q'\) in terms of \(\alpha\)?