a. [5 marks]

(i) \(\overrightarrow{OA} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\), \(\overrightarrow{OB} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix}\).
\(\overrightarrow{OA} \times \overrightarrow{OB} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 3 & 1 & 2 \end{vmatrix} = \mathbf{i}(2 \cdot 2 – 3 \cdot 1) – \mathbf{j}(1 \cdot 2 – 3 \cdot 3) + \mathbf{k}(1 \cdot 1 – 2 \cdot 3) = \mathbf{i}(4 – 3) – \mathbf{j}(2 – 9) + \mathbf{k}(1 – 6) = \mathbf{i} + 7\mathbf{j} – 5\mathbf{k}\) (A1).
(ii) Area of triangle \( OAB = \frac{1}{2} |\overrightarrow{OA} \times \overrightarrow{OB}| = \frac{1}{2} \sqrt{1^2 + 7^2 + (-5)^2} = \frac{1}{2} \sqrt{1 + 49 + 25} = \frac{1}{2} \sqrt{75} = \frac{5\sqrt{3}}{2} \approx 4.33\) (M1A1).
(iii) Normal to plane \( OAB \) is \(\overrightarrow{OA} \times \overrightarrow{OB} = \begin{pmatrix} 1 \\ 7 \\ -5 \end{pmatrix}\). Equation: \(\mathbf{r} \cdot \begin{pmatrix} 1 \\ 7 \\ -5 \end{pmatrix} = k\) (M1).
Use point \( O(0, 0, 0) \): \(0 + 7 \cdot 0 – 5 \cdot 0 = k \Rightarrow k = 0\).
Cartesian equation: \( x + 7y – 5z = 0 \) (A1).

b. [7 marks]

(i) Direction of \( L_1 \): \(\overrightarrow{AB} = \begin{pmatrix} 3 – 1 \\ 1 – 2 \\ 2 – 3 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix}\) (M1A1).
Vector equation of \( L_1 \): \(\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix}\) (A1).
(ii) For intersection, set \(\begin{pmatrix} 1 + 2\lambda \\ 2 – \lambda \\ 3 – \lambda \end{pmatrix} = \begin{pmatrix} 2 + \mu \\ 4 + 3\mu \\ 3 + 2\mu \end{pmatrix}\):
\(1 + 2\lambda = 2 + \mu\) (1)
\(2 – \lambda = 4 + 3\mu\) (2) (M1A1)
\(3 – \lambda = 3 + 2\mu\) (3)
From (2): \(\lambda = -4 – 3\mu\). Substitute into (3): \(3 – (-4 – 3\mu) = 3 + 2\mu \Rightarrow 7 + 3\mu = 3 + 2\mu \Rightarrow \mu = -4\).
Then: \(\lambda = -4 – 3(-4) = 8\). Check (1): \(1 + 2(8) = 17 \neq 2 – 4 = -2\).
Values do not satisfy (1), so lines are skew (A1R1).