The line \(L _ { 1 }\) has equation \(\mathbf { r } = \left( \begin{array} { c } - 13 7 - 1 \end{array} \right) + t \left( \begin{array} { c } 6 - 2 3 \end{array} \right)\). The line \(L _ { 2 }\) passes through the point \(A\) with position vector \(\left( \begin{array} { c } 1 p 10 \end{array} \right)\) and is parallel to \(\left( \begin{array} { c } - 2 11 - 5 \end{array} \right)\), where \(p\)
is a constant. The lines \(L _ { 1 }\) and \(L _ { 2 }\) intersect at the point \(B\).
Find
the value of \(p\),
the position vector of \(B\).
The point \(C\) lies on \(L _ { 1 }\) and angle \(A C B\) is \(90 ^ { \circ }\)
Find the position vector of \(C\).
The point \(D\) also lies on \(L _ { 1 }\) and triangle \(A B D\) is isosceles with \(A B = A D\).