(a)
Condition: Lines perpendicular, dot product of direction vectors = 0
Direction vectors: \(L_1: \begin{pmatrix} p \\ 2p \\ 4 \end{pmatrix}\), \(L_2: \begin{pmatrix} p + 4 \\ 4 \\ -7 \end{pmatrix}\)
Dot product: \(p \times (p + 4) + (2p) \times 4 + 4 \times (-7) = p^2 + 4p + 8p – 28 = p^2 + 12p – 28\)
Equation: \(p^2 + 12p – 28 = 0\)
Solve: \(p = \frac{-12 \pm \sqrt{256}}{2} = 2, -14\)
Marks: (M1) for dot product condition, (A1) for vectors, (A1) for dot product, (A1) for solutions
Result: \(p = 2, -14\) [4]

(b)
Use \(p = -14\) (since \(p < 0\))
Lines: \(L_1: \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} + \lambda \begin{pmatrix} -14 \\ -28 \\ 4 \end{pmatrix}\), \(L_2: \begin{pmatrix} 14 \\ 7 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} -10 \\ 4 \\ -7 \end{pmatrix}\)
Equate: \(2 – 14\lambda = 14 – 10\mu\), \(-5 – 28\lambda = 7 + 4\mu\), \(-3 + 4\lambda = -2 – 7\mu\)
Solve: From (i) \(14\lambda – 10\mu = -12\), (ii) \(28\lambda + 4\mu = -12\)
Then: \(28\lambda – 20\mu = -24\), so \(-24\mu = -12\), \(\mu = \frac{1}{2}\), \(\lambda = -\frac{1}{2}\)
Check (iii): \(-3 + 4 \times (-\frac{1}{2}) = -5\), \(-2 – 7 \times \frac{1}{2} = -5.5\), \(-5 \neq -5.5\), no solution
Marks: (M1) for choosing \(p\), (A1) for equations, (A1) for conclusion
Result: Lines do not intersect [3]