3 You are given the matrix \(\mathbf { A } = \left( \begin{array} { r r r } k & - 7 & 4
2 & - 2 & 3
1 & - 3 & - 2 \end{array} \right)\).

1. Show that when \(k = 5\) the determinant of \(\mathbf { A }\) is zero. Obtain an expression for the inverse of \(\mathbf { A }\) when \(k \neq 5\).
2. Solve the matrix equation $$\left( \begin{array} { r r r } 4 & - 7 & 4
   2 & - 2 & 3
   1 & - 3 & - 2 \end{array} \right) \left( \begin{array} { l } x
   y
   z \end{array} \right) = \left( \begin{array} { c } p
   1
   2 \end{array} \right)$$ giving your answer in terms of \(p\).
3. Find the value of \(p\) for which the matrix equation $$\left( \begin{array} { r r r } 5 & - 7 & 4
   2 & - 2 & 3
   1 & - 3 & - 2 \end{array} \right) \left( \begin{array} { c } x
   y
   z \end{array} \right) = \left( \begin{array} { c } p
   1
   2 \end{array} \right)$$ has a solution. Give the general solution in this case and describe it geometrically.