a. [5 marks]

Set \(\mathbf{r}_1 = \mathbf{r}_2\): \(\begin{pmatrix} 1 + 2s \\ 2 + 3s \\ -3 + 6s \end{pmatrix} = \begin{pmatrix} 9 + 4t \\ 9 + t \\ 11 + 2t \end{pmatrix}\) (M1).
Equate components:
\(1 + 2s = 9 + 4t\) (A1) (1)
\(2 + 3s = 9 + t\) (A1) (2)
\(-3 + 6s = 11 + 2t\) (A1) (3)
From (2): \(t = 3s – 7\).
Substitute into (1): \(1 + 2s = 9 + 4(3s – 7) = 12s – 19 \Rightarrow -10s = -20 \Rightarrow s = 2\) (M1).
Then: \(t = 3(2) – 7 = -1\).
Verify with (3): \(-3 + 6(2) = 11 + 2(-1) = 9\), satisfied (A1).
Position vector of \(A\): \(\begin{pmatrix} 1 + 2(2) \\ 2 + 3(2) \\ -3 + 6(2) \end{pmatrix} = \begin{pmatrix} 5 \\ 8 \\ 9 \end{pmatrix}\) (A1).

b. [3 marks]

For \(L\): \(\begin{pmatrix} 1 + 2s \\ 2 + 3s \\ -3 + 6s \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} = 2(2 + 3s) – 1(-3 + 6s) = 4 + 6s + 3 – 6s = 7\) (M1A1).
For \(M\): \(\begin{pmatrix} 9 + 4t \\ 9 + t \\ 11 + 2t \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} = 2(9 + t) – 1(11 + 2t) = 18 + 2t – 11 – 2t = 7\) (A1).
Both lines satisfy the plane equation, so they lie in \(\Pi\).

c(i). [4 marks]

Line through \(B(-3, 12, 2)\) perpendicular to \(\Pi\) has direction vector \(\begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix}\) (A1).
Parametric equation: \(\mathbf{r} = \begin{pmatrix} -3 \\ 12 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix}\) (M1).
Substitute into \(\mathbf{r} \cdot \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} = 7\):
\(\begin{pmatrix} -3 \\ 12 + 2\lambda \\ 2 – \lambda \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} = 2(12 + 2\lambda) – (2 – \lambda) = 24 + 4\lambda – 2 + \lambda = 22 + 5\lambda = 7 \Rightarrow 5\lambda = -15 \Rightarrow \lambda = -3\) (A1).
Position vector of \(C\): \(\begin{pmatrix} -3 \\ 12 + 2(-3) \\ 2 – (-3) \end{pmatrix} = \begin{pmatrix} -3 \\ 6 \\ 5 \end{pmatrix}\) (A1).

c(ii). [3 marks]

\(\overrightarrow{BC} = \begin{pmatrix} -3 – (-3) \\ 6 – 12 \\ 5 – 2 \end{pmatrix} = \begin{pmatrix} 0 \\ -6 \\ 3 \end{pmatrix}\) (M1).
\(|\overrightarrow{BC}| = \sqrt{0^2 + (-6)^2 + 3^2} = \sqrt{45} = 3\sqrt{5} \approx 6.71\) (A1A1).

d. [3 marks]

Reflection \(B’\) has \(C\) as the midpoint of \(B\) and \(B’\).
\(\mathbf{r}_{B’} = \mathbf{r}_C + (\mathbf{r}_C – \mathbf{r}_B)\) (M1).
\(\mathbf{r}_C – \mathbf{r}_B = \begin{pmatrix} -3 – (-3) \\ 6 – 12 \\ 5 – 2 \end{pmatrix} = \begin{pmatrix} 0 \\ -6 \\ 3 \end{pmatrix}\) (A1).
\(\mathbf{r}_{B’} = \begin{pmatrix} -3 \\ 6 \\ 5 \end{pmatrix} + \begin{pmatrix} 0 \\ -6 \\ 3 \end{pmatrix} = \begin{pmatrix} -3 \\ 0 \\ 8 \end{pmatrix}\) (A1).