14. Relative to a fixed origin \(O\), the line \(l\) has vector equation
$$\mathbf { r } = \left( \begin{array} { r }
- 1
- 4
6
\end{array} \right) + \lambda \left( \begin{array} { r }
2
1
- 1
\end{array} \right)$$
where \(\lambda\) is a scalar parameter.
Points \(A\) and \(B\) lie on the line \(l\), where \(A\) has coordinates ( \(1 , a , 5\) ) and \(B\) has coordinates ( \(b , - 1,3\) ).
- Find the value of the constant \(a\) and the value of the constant \(b\).
- Find the vector \(\overrightarrow { A B }\).
The point \(C\) has coordinates ( \(4 , - 3,2\) )
- Show that the size of the angle \(C A B\) is \(30 ^ { \circ }\)
- Find the exact area of the triangle \(C A B\), giving your answer in the form \(k \sqrt { 3 }\), where \(k\) is a constant to be determined.
The point \(D\) lies on the line \(l\) so that the area of the triangle \(C A D\) is twice the area of the triangle \(C A B\).
- Find the coordinates of the two possible positions of \(D\).