a. [4 marks]

Attempting to find a normal to \(\pi\), e.g., \(\begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} \times \begin{pmatrix} 8 \\ 11 \\ 6 \end{pmatrix}\) (M1).
\(\begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} \times \begin{pmatrix} 8 \\ 11 \\ 6 \end{pmatrix} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 2 & -2 \\ 8 & 11 & 6 \end{vmatrix} = \begin{pmatrix} (2)(6) – (-2)(11) \\ -((3)(6) – (-2)(8)) \\ (3)(11) – (2)(8) \end{pmatrix} = \begin{pmatrix} 34 \\ -34 \\ 17 \end{pmatrix} = 17 \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}\) (A1).
\(\mathbf{r} \cdot \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \\ 12 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} = 2(1) + (-2)(5) + 1(12) = 2 – 10 + 12 = 4\) (M1).
\(2x – 2y + z = 4\) (or equivalent, e.g., \(x – y + \frac{1}{2}z = 2\)) (A1).

b. [4 marks]

\(l_3: \mathbf{r} = \begin{pmatrix} 4 \\ 0 \\ 8 \end{pmatrix} + t \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}, t \in \mathbb{R}\) (A1).
Attempting to solve \(\begin{pmatrix} 4 + 2t \\ -2t \\ 8 + t \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} = 4\), i.e., \(2(4 + 2t) + (-2)(-2t) + 1(8 + t) = 8 + 4t + 4t + 8 + t = 9t + 16 = 4\) for \(t\) (M1).
\(9t + 16 = 4 \Rightarrow 9t = -12 \Rightarrow t = -\frac{4}{3}\) (A1).
Coordinates: \(\begin{pmatrix} 4 + 2\left(-\frac{4}{3}\right) \\ -2\left(-\frac{4}{3}\right) \\ 8 + \left(-\frac{4}{3}\right) \end{pmatrix} = \begin{pmatrix} 4 – \frac{8}{3} \\ \frac{8}{3} \\ 8 – \frac{4}{3} \end{pmatrix} = \begin{pmatrix} \frac{4}{3} \\ \frac{8}{3} \\ \frac{20}{3} \end{pmatrix}\) (A1).