(a) To find \(CAB\), first compute \(AB\):

\(AB = \begin{pmatrix} 2 & k & k \\ 5 & -1 & 3 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 2+k & k \\ 8 & -1 \\ 2 & 0 \end{pmatrix}\)

Then compute \(CAB\):

\(CAB = \begin{pmatrix} 0 & 1 & 1 \\ -1 & 2 & 0 \end{pmatrix} \begin{pmatrix} 2+k & k \\ 8 & -1 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} 10 & -1 \\ -k+14 & -k-2 \end{pmatrix}\)

(b) Since \(A\) is singular, its determinant is zero:

\(\det(A) = 2(-1 \cdot 1 - 3 \cdot 0) - k(5 \cdot 1 - 3 \cdot 1) + k(5 \cdot 0 - (-1) \cdot 1) = 0\)

\(-2 - 2k + k = 0\)

\(-2 - k = 0\)

\(k = -2\)

(c) Using \(k = -2\), the matrix \(CAB\) becomes:

\(\begin{pmatrix} 10 & -1 \\ 16 & 0 \end{pmatrix}\)

Transform \(\begin{pmatrix} x \\ y \end{pmatrix}\):

\(\begin{pmatrix} 10 & -1 \\ 16 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 10x - y \\ 16x \end{pmatrix}\)

Equating \(y = mx\) and \(Y = mX\):

\(10x - mx = X\)

\(16x = mX\)

\(16 = 10m - m^2\)

\(m^2 - 10m + 16 = 0\)

Solving gives \(m = 2\) and \(m = 8\), so the invariant lines are:

\(y = 2x\) and \(y = 8x\)