(a) The polar equation is \(r = a \sec^2 \theta\). At \(\theta = 0\), \(r = a \sec^2(0) = a\), giving the point \((a, 0)\). At \(\theta = \frac{1}{4} \pi\), \(r = a \sec^2(\frac{1}{4} \pi) = 2a\), giving the point \((2a, \frac{1}{4} \pi)\).

(b) The maximum distance from the initial line is when \(\theta = \frac{1}{4} \pi\), where \(r = 2a\). The distance is \(y = r \sin \theta = 2a \sin(\frac{1}{4} \pi) = a\sqrt{2}\).

(c) The area \(A\) is given by \(\frac{1}{2} \int_0^{\frac{1}{4} \pi} r^2 \, d\theta\). Substituting \(r = a \sec^2 \theta\), we have:

\(A = \frac{1}{2} a^2 \int_0^{\frac{1}{4} \pi} \sec^4 \theta \, d\theta\)

\(= \frac{1}{2} a^2 \int_0^{\frac{1}{4} \pi} (\sec^2 \theta)(\tan^2 \theta + 1) \, d\theta\)

\(= \frac{1}{2} a^2 \left[ \tan \theta + \frac{1}{3} \tan^3 \theta \right]_0^{\frac{1}{4} \pi}\)

\(= \frac{3}{8} a^2\)

(d) The Cartesian form is derived by substituting \(x = r \cos \theta\) and \(y = r \sin \theta\). From \(r = a \sec^2 \theta\), we have:

\(x^2 + y^2 = r^2\)

\(r^2 \cos^2 \theta = a^2 \Rightarrow x^2 = a^2 - y^2\)

\(x^4 = a^2 (x^2 + y^2)\)

\(y = \sqrt{a^2 x^4 - x^2} = x \sqrt{a^2 x^2 - 1}\)

(a) To find the length \(PQ\), we first find a vector joining any point on \(l_1\) to any point on \(l_2\). The vector is \(\begin{pmatrix} 0 \\ 2 \\ 6 \end{pmatrix} - \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 2 \\ 5 \end{pmatrix}\).

Next, find the common perpendicular using the cross product: \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ 1 & 2 & -2 \end{vmatrix} = \begin{pmatrix} 2 \\ -4 \\ -3 \end{pmatrix}\).

The length \(PQ\) is given by \(\frac{1}{\sqrt{29}} \begin{pmatrix} -2 \\ 2 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -4 \\ -3 \end{pmatrix} = \frac{27}{\sqrt{29}} = 5.01\).

(b)(i) The equation of \(\Pi_1\) is \(\mathbf{r} = 2\mathbf{i} + \mathbf{k} + s(\mathbf{i} - \mathbf{j} + 2\mathbf{k}) + t(2\mathbf{i} - 4\mathbf{j} - 3\mathbf{k})\).

(b)(ii) The normal to \(\Pi_2\) is found using the cross product: \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -2 \\ 2 & -4 & -3 \end{vmatrix} = \begin{pmatrix} -14 \\ -1 \\ -8 \end{pmatrix}\).

Substitute a point to find the equation: \(14(0) + (2) + 8(6) = 50 \Rightarrow 14x + y + 8z = 50\).

(c) The normals to \(\Pi_1\) and \(\Pi_2\) are \(\begin{pmatrix} 11 \\ 7 \\ -2 \end{pmatrix}\) and \(\begin{pmatrix} 14 \\ 1 \\ 8 \end{pmatrix}\) respectively.

Use the dot product to find the angle: \(\cos \theta = \frac{145}{\sqrt{174} \sqrt{261}}\).

The acute angle is \(47.1^\circ\).

(a) The vertical asymptote occurs where the denominator is zero: \(x + 1 = 0 \Rightarrow x = -1\).
The oblique asymptote is found by dividing \(x^2 - x\) by \(x + 1\), giving \(y = x - 2\).

(b) Differentiate \(y = \frac{x^2 - x}{x + 1}\) using the quotient rule:
\(\frac{dy}{dx} = \frac{(2x - 1)(x + 1) - (x^2 - x)}{(x + 1)^2} = \frac{1 - 2(x + 1)^2}{(x + 1)^2}\).
Set \(\frac{dy}{dx} = 0\) to find stationary points:
\(1 - 2(x + 1)^2 = 0 \Rightarrow (x + 1)^2 = \frac{1}{2}\).
Solving gives \(x = -1 \pm \sqrt{2}\).
Substitute back to find \(y\):
\(y = \frac{(-1 + \sqrt{2})^2 - (-1 + \sqrt{2})}{-1 + \sqrt{2} + 1} = -3 + 2\sqrt{2}\),
\(y = \frac{(-1 - \sqrt{2})^2 - (-1 - \sqrt{2})}{-1 - \sqrt{2} + 1} = -3 - 2\sqrt{2}\).

(c) The sketch shows the curve with asymptotes and intersections at \((0, 0)\) and \((1, 0)\).

(d) Solve \(\left| \frac{x^2 - x}{x + 1} \right| < 6\):
\(\frac{x^2 - x}{x + 1} < 6\) or \(\frac{x^2 - x}{x + 1} > -6\).
For \(\frac{x^2 - x}{x + 1} < 6\):
\(x^2 - 7x - 6 = 0 \Rightarrow x = -3, -2\).
For \(\frac{x^2 - x}{x + 1} > -6\):
\(x^2 + 5x + 6 = 0 \Rightarrow x = -2, -3\).
Thus, \(-3 < x < -2 \text{ and } \frac{1}{2} - \frac{1}{2}\sqrt{73} < x < \frac{1}{2} + \frac{1}{2}\sqrt{73}\).