(a) Express the general point of at least one line in component form: \((1 + a\lambda, 2 + 2\lambda, 1 - \lambda)\) or \((2 + 2\mu, 1 - \mu, -1 + \mu)\).

Equate at least two pairs of corresponding components and solve for \(\lambda\) or \(\mu\):

\(1 + a\lambda = 2 + 2\mu\)

\(2 + 2\lambda = 1 - \mu\)

\(1 - \lambda = -1 + \mu\)

Solving gives \(\lambda = -3\) or \(\mu = 5\).

Obtain \(a = \frac{11}{3}\).

The point of intersection has position vector \(12\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}\).

(b) Use the correct process for finding the scalar product of direction vectors for the two lines: \((a, 2, -1)\) and \((2, -1, 1)\).

Using the correct process for the moduli, divide the scalar product by the product of the moduli and equate the result to \(\frac{1}{6}\).

State a correct equation in \(a\) in any form, e.g., \(\frac{2a - 2 - 1}{\sqrt{6(a^2 + 5)}} = \pm \frac{1}{6}\).

Solve for \(a\):

\(36(2a - 3)^2 = 6a^2 + 5\)

\(23a^2 - 72a + 49 = 0\)

Obtain \(a = 1\) or \(a = \frac{49}{23}\).