(a) Since \(\mathbf{A}\) is singular, its determinant is zero. Calculate the determinant:

\(\det(\mathbf{A}) = 1(1 \cdot 5 - 3 \cdot 2) - k(2 \cdot 5 - 3 \cdot 3) + 3(2 \cdot 2 - 1 \cdot 3) = 0.\)

This simplifies to \(-1 - k + 3 = 0 \Rightarrow k = 2.\)

Substitute \(k = 2\) into \(\mathbf{A}\) and compute \(\mathbf{CAB}\):

\(\mathbf{C} \cdot \mathbf{A} = \begin{pmatrix} -2 & -1 \\ 1 & 1 \\ 1 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 2 & 5 \end{pmatrix} = \begin{pmatrix} -2 & -4 & -11 \\ 1 & 3 & 6 \\ 1 & 5 & 12 \end{pmatrix}.\)

Then, \(\mathbf{CAB} = \begin{pmatrix} -2 & -4 & -11 \\ 1 & 3 & 6 \end{pmatrix} \cdot \begin{pmatrix} 0 & -2 \\ -1 & 3 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix}.\)

(b) For invariant lines, solve \(\begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = m \begin{pmatrix} x \\ y \end{pmatrix}.\)

This gives \(-9x + 3mx = m(3x - 7mx)\) and \(-9 + 3m = 3m - 7m^2 \Rightarrow 7m^2 = 9.\)

Thus, \(m = \pm \frac{3}{\sqrt{7}}\), giving lines \(y = \frac{3}{\sqrt{7}}x\) and \(y = -\frac{3}{\sqrt{7}}x\).

(c) Assume \(\mathbf{D} = \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix}, \mathbf{E} = \begin{pmatrix} \beta & 0 \\ 0 & 1 \end{pmatrix}, \mathbf{F} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.\)

Given \(\mathbf{CAB} = \mathbf{D} - 9\mathbf{EF},\) solve:

\(\begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} = \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} - 9 \begin{pmatrix} 0 & \beta \\ 1 & 0 \end{pmatrix}.\)

Equating gives \(\alpha = 3\) and \(\beta = \frac{7}{9}.\)

Thus, \(\mathbf{D} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}, \mathbf{E} = \begin{pmatrix} \frac{7}{9} & 0 \\ 0 & 1 \end{pmatrix}, \mathbf{F} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.\)