6 The line \(l _ { 1 }\) has equation \(\mathbf { r } = \left[ \begin{array} { r } 4
- 5
3 \end{array} \right] + \lambda \left[ \begin{array} { r } - 1
3
1 \end{array} \right]\).
The line \(l _ { 2 }\) has equation \(\mathbf { r } = \left[ \begin{array} { r } 7
- 8
6 \end{array} \right] + \mu \left[ \begin{array} { r } 2
- 3
1 \end{array} \right]\).
The point \(P\) lies on \(l _ { 1 }\) where \(\lambda = - 1\). The point \(Q\) lies on \(l _ { 2 }\) where \(\mu = 2\).

1. Show that the vector \(\overrightarrow { P Q }\) is parallel to \(\left[ \begin{array} { r } 1
   - 1
   1 \end{array} \right]\).
2. The lines \(l _ { 1 }\) and \(l _ { 2 }\) intersect at the point \(R ( 3 , b , c )\).
   1. Show that \(b = - 2\) and find the value of \(c\).
   2. The point \(S\) lies on a line through \(P\) that is parallel to \(l _ { 2 }\). The line \(R S\) is perpendicular to the line \(P Q\).
      \includegraphics[max width=\textwidth, alt={}, center]{9f03a5f3-7fea-4fb7-b3bd-b4c0cdf662a2-16_887_1159_1320_443} Find the coordinates of \(S\).
      \(7 \quad\) A curve has equation \(\cos 2 y + y \mathrm { e } ^ { 3 x } = 2 \pi\).
      The point \(A \left( \ln 2 , \frac { \pi } { 4 } \right)\) lies on this curve.