(a) Express the general point of l or m in component form:

\((-1 + 2\lambda, 3 - \lambda, 4 - \lambda)\)

\((5 + a\mu, 4 + b\mu, 3 + \mu)\)

Equate components and eliminate either \(\lambda\) or \(\mu\):

\(\mu = \frac{2\lambda - 6}{a}\)

\(\mu = \frac{-1 - \lambda}{b}\)

Use \(\lambda = 1 - \mu\) to obtain \(2b - a = 4\).

(b) For perpendicular lines, the scalar product of direction vectors is zero:

\((2\mathbf{i} - \mathbf{j} - \mathbf{k}) \cdot (a\mathbf{i} + b\mathbf{j} + \mathbf{k}) = 0\)

\(2a - b - 1 = 0\)

Solve simultaneous equations:

\(2a - b = 1\)

\(2b - a = 4\)

Obtain \(a = 2, b = 3\).

(c) Substitute \(a = 2\) and \(b = 3\) into component equations and solve for \(\lambda\) or \(\mu\):

Obtain position vector \(3\mathbf{i} + \mathbf{j} + 2\mathbf{k}\).