(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 \(y = 8x\)

(a) The curve is approximately correct, passing through the pole, \(O\), with the correct domain. The curve \(r\) is strictly increasing over the domain \(0\) to \(\frac{\pi}{2}\). It has the correct form at \(O\) and is approximately tangential to the initial line at \(O\).

(b) To find the area, use the formula:

\(\frac{1}{2} \int_0^{\frac{1}{2}\pi} \left( \frac{1}{\pi - \theta} - \frac{1}{\pi} \right)^2 \, d\theta\)

Expanding the integrand:

\(\frac{1}{2} \int_0^{\frac{1}{2}\pi} \left( \frac{1}{(\pi - \theta)^2} - \frac{2}{\pi(\pi - \theta)} + \frac{1}{\pi^2} \right) \, d\theta\)

Integrating each term separately:

\(\frac{1}{2} \left[ \frac{1}{\pi - \theta} + \frac{2}{\pi} \ln(\pi - \theta) + \frac{\theta}{\pi^2} \right]_0^{\frac{1}{2}\pi}\)

Substituting the limits:

\(\frac{1}{2} \left( \frac{2}{\pi} - \frac{\pi}{2} + \frac{2}{\pi} \ln \frac{\pi}{2} + \frac{1}{2\pi} \right) = \frac{1}{2} \left( \frac{3}{2\pi} + \frac{2}{\pi} \ln \frac{1}{2} \right)\)

Simplifying the logarithmic terms:

\(= \frac{3 - 4 \ln 2}{4\pi}\)

(a) The direction vector of \(l_1\) is \(2\mathbf{i} - 3\mathbf{j}\). The vector from \(P\) to a point on \(l_1\) is \(\mathbf{i} - 3\mathbf{k}\). Thus, the equation of \(\Pi_1\) is \(\mathbf{r} = -2\mathbf{i} - 2\mathbf{j} + 4\mathbf{k} + \lambda (2\mathbf{i} - 3\mathbf{j}) + \mu (\mathbf{i} - 3\mathbf{k})\).

(b) The normal vector to \(\Pi_2\) is \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 0 \\ 3 & -1 & 3 \end{vmatrix} = -9\mathbf{i} - 6\mathbf{j} + 7\mathbf{k}\). Using the point \((3, 0, -2)\) on \(\Pi_2\), the equation is \(9x + 6y - 7z = 41\).

(c) The normal vectors to \(\Pi_1\) and \(\Pi_2\) are \(\begin{pmatrix} 9 \\ 6 \\ -7 \end{pmatrix}\) and \(\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}\) respectively. The dot product is \(32\), and the magnitudes are \(\sqrt{166}\) and \(\sqrt{14}\). Thus, \(\cos \alpha = \frac{32}{\sqrt{14 \times 166}}\), giving \(\alpha = 48.4^\circ\) or \(0.845 \text{ rad}\).

(d) \(\overrightarrow{OQ} = -5\overrightarrow{OP} = \begin{pmatrix} 10 \\ 10 \\ -20 \end{pmatrix}\). The equation of \(\Pi_2\) is \(9x + 6y - 7z = 41\). Substituting \(\mathbf{OF} = \begin{pmatrix} 10 + 9t \\ 10 + 6t \\ -20 - 7t \end{pmatrix}\) into the plane equation gives \(t = -\frac{3}{2}\), leading to \(\mathbf{OF} = \left( -\frac{7}{2}, 1, -\frac{19}{2} \right)\).

(a) The vertical asymptotes occur where the denominator is zero: \(1 + x - x^2 = 0\). Solving \(x^2 - x - 1 = 0\) gives \(x = \frac{1 \pm \sqrt{5}}{2}\). The horizontal asymptote is found by considering the limits as \(x \to \pm \infty\), giving \(y = -1\).

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Using the quotient rule, \(\frac{dy}{dx} = \frac{(1 + x - x^2)(2x - 1) - (x^2 - x - 3)(1 - 2x)}{(1 + x - x^2)^2}\). Simplifying gives \((2x - 1)(x - 2) = 0\), so \(x = \frac{1}{2}\). Substituting back, \(y = -\frac{13}{9}\), giving the point \(\left( \frac{1}{2}, -\frac{13}{9} \right)\).

(c) The intersections with the axes are found by setting \(y = 0\) and \(x = 0\). Solving \(x^2 - x - 3 = 0\) gives \(x = \frac{1}{2} \pm \frac{1}{2}\sqrt{13}\). For \(x = 0\), \(y = -3\). Thus, intersections are \(\left( \frac{1}{2} + \frac{1}{2}\sqrt{13}, 0 \right), \left( \frac{1}{2} - \frac{1}{2}\sqrt{13}, 0 \right), (0, -3)\).

(d) For \(\left| \frac{x^2 - x - 3}{1 + x - x^2} \right| < 3\), solve \(\frac{x^2 - x - 3}{1 + x - x^2} < 3\) and \(\frac{x^2 - x - 3}{1 + x - x^2} > -3\). Solving gives critical points \(x = \frac{1}{2} \pm \frac{1}{2}\sqrt{7}, x = 0, x = 1\). The solution is \(x < \frac{1}{2} - \frac{1}{2}\sqrt{7}, \; 0 < x < 1, \; x > \frac{1}{2} + \frac{1}{2}\sqrt{7}\).

Given \(\alpha + \beta + \gamma = 3\), we have \(b = - (\alpha + \beta + \gamma) = -3\).

Using the formula for the sum of squares: \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\),

\(5 = 3^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\).

Solving gives \(\alpha\beta + \beta\gamma + \gamma\alpha = 1\), so \(c = 2\).

Using the formula for the sum of cubes: \(\alpha^3 + \beta^3 + \gamma^3 = 3(\alpha\beta\gamma) + (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha)\),

\(6 = 3(\alpha\beta\gamma) + 3(5 - 1)\).

Solving gives \(\alpha\beta\gamma = 1\), so \(d = 1\).

The equation is \(x^3 - 3x^2 + 2x + 1 = 0\).