1. The line \(l _ { 1 }\) has vector equation

$$\mathbf { r } = \left( \begin{array} { l } 3
1
2 \end{array} \right) + \lambda \left( \begin{array} { r } 1
- 1
4 \end{array} \right)$$ and the line \(l _ { 2 }\) has vector equation $$\mathbf { r } = \left( \begin{array} { r } 0
4
- 2 \end{array} \right) + \mu \left( \begin{array} { r } 1
- 1
0 \end{array} \right) ,$$ where \(\lambda\) and \(\mu\) are parameters.
The lines \(l _ { 1 }\) and \(l _ { 2 }\) intersect at the point \(B\) and the acute angle between \(l _ { 1 }\) and \(l _ { 2 }\) is \(\theta\).

1. Find the coordinates of \(B\).
2. Find the value of \(\cos \theta\), giving your answer as a simplified fraction. The point \(A\), which lies on \(l _ { 1 }\), has position vector \(\mathbf { a } = 3 \mathbf { i } + \mathbf { j } + 2 \mathbf { k }\).
   The point \(C\), which lies on \(l _ { 2 }\), has position vector \(\mathbf { c } = 5 \mathbf { i } - \mathbf { j } - 2 \mathbf { k }\).
   The point \(D\) is such that \(A B C D\) is a parallelogram.
3. Show that \(| \overrightarrow { A B } | = | \overrightarrow { B C } |\).
4. Find the position vector of the point \(D\).