(a) The curve \(C\) has polar equation \(r = 3 + 2\sin\theta\) for \(-\pi < \theta \le \pi\). It is a limacon, symmetric about the line \(\theta = \frac{\pi}{2}\); the completed sketch shows a single loop above the pole.

(b) The line \(l\) has polar equation \[ r\sin\theta = 2 \quad\Rightarrow\quad r = 2\csc\theta \quad (\sin\theta \ne 0). \] Points of intersection satisfy \[ 3 + 2\sin\theta \;=\; 2\csc\theta. \] Multiplying by \(\sin\theta\), \[ 3\sin\theta + 2\sin^2\theta = 2. \] Let \(\sin\theta = s\). Then \[ 2s^2 + 3s - 2 = 0 \quad\Rightarrow\quad s = \frac{-3 \pm 5}{4} \;\Rightarrow\; s = \frac12 \text{ or } s = -2. \] \(\sin\theta = -2\) is impossible, so \(\sin\theta = \frac12\), giving \[ \theta = \frac{\pi}{6},\; \frac{5\pi}{6}. \] At these points, \[ r = 3 + 2\sin\theta = 3 + 2\cdot\frac12 = 4. \] So the points of intersection are \[ \left(4,\; \frac{\pi}{6}\right),\quad \left(4,\; \frac{5\pi}{6}\right). \]

(c) The region \(R\) is enclosed by \(C\) and the line \(l\), and contains the pole. For a given \(\theta\), the boundary closest to the pole is:

• \(r = 3 + 2\sin\theta\) for \(-\pi < \theta \le \frac{\pi}{6}\) and \(\frac{5\pi}{6} \le \theta \le \pi\) (the curve is nearer than the line),
• \(r = 2\csc\theta\) for \(\frac{\pi}{6} \le \theta \le \frac{5\pi}{6}\) (the line is nearer than the curve).

Hence the area of \(R\) is \[ \text{Area}(R) = \frac12 \int_{-\pi}^{\pi/6} (3 + 2\sin\theta)^2\,d\theta + \frac12 \int_{\pi/6}^{5\pi/6} (2\csc\theta)^2\,d\theta + \frac12 \int_{5\pi/6}^{\pi} (3 + 2\sin\theta)^2\,d\theta. \] It is convenient to rewrite this as \[ \text{Area}(R) = \frac12 \int_{-\pi}^{\pi} (3 + 2\sin\theta)^2\,d\theta \;-\; \frac12 \int_{\pi/6}^{5\pi/6} \Bigl[(3 + 2\sin\theta)^2 - (2\csc\theta)^2\Bigr]\,d\theta. \]

First, \[ \int_{-\pi}^{\pi} (3 + 2\sin\theta)^2\,d\theta = \int_{-\pi}^{\pi} \bigl(9 + 12\sin\theta + 4\sin^2\theta\bigr)\,d\theta = 22\pi. \]

Next, \[ (3 + 2\sin\theta)^2 - (2\csc\theta)^2 = 9 + 12\sin\theta + 4\sin^2\theta - \frac{4}{\sin^2\theta}, \] so \[ \int_{\pi/6}^{5\pi/6} \Bigl[(3 + 2\sin\theta)^2 - (2\csc\theta)^2\Bigr]\,d\theta = 5\sqrt{3} + \frac{22\pi}{3}. \]

Therefore, \[ \text{Area}(R) = \frac12\left(22\pi - \left(5\sqrt{3} + \frac{22\pi}{3}\right)\right) = \frac12\left(\frac{44\pi}{3} - 5\sqrt{3}\right) = \frac{22}{3}\pi - \frac{5}{2}\sqrt{3}. \]

So the exact area of \(R\) is \[ \boxed{\displaystyle \frac{22}{3}\pi - \frac{5}{2}\sqrt{3}}. \]

(a) To find the vertical asymptote, set the denominator equal to zero: \(x - 3 = 0\), so \(x = 3\).
For the oblique asymptote, perform polynomial long division on \(\frac{x^2}{x-3}\):

\(x^2 \div (x-3) = x + 3 + \frac{9}{x-3}\).
As \(x \to \infty\), \(\frac{9}{x-3} \to 0\), so the oblique asymptote is \(y = x + 3\).

(b) Substitute \(y = \frac{x^2}{x-3}\) into \(yx - 3y = x^2\) to form a quadratic: \(x^2 - yx + 3y = 0\).
The discriminant \(b^2 - 4ac = y^2 - 4(3y) = y^2 - 12y\) must be negative for no real solutions.
\(y^2 - 12y < 0\) implies \(0 < y < 12\), showing no points exist for \(0 < y < 12\).

(c) Sketch the curve showing the vertical asymptote at \(x = 3\) and the oblique asymptote \(y = x + 3\). The curve approaches these asymptotes.

(d)(i) Sketch \(y = \left| \frac{x^2}{x-3} \right|\) and \(y = |x| - 3\).
Find intercepts: \(y = \left| \frac{x^2}{x-3} \right|\) intersects the y-axis at \((0,0)\).
\(y = |x| - 3\) intersects the y-axis at \((0,-3)\) and x-axis at \((3,0)\) and \((-3,0)\).

(d)(ii) From the sketch, \(\left| \frac{x^2}{x-3} \right| \leq |x| + c\) has no solution when \(c < -3\).

(a) To find the equation of the plane OAB, we need a normal vector. The vectors \(\overrightarrow{OA} = \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix}\) and \(\overrightarrow{OB} = \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}\) are in the plane. The cross product \(\overrightarrow{OA} \times \overrightarrow{OB}\) gives the normal vector \(\mathbf{n} = \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\). The equation of the plane is \(\mathbf{r} \cdot \mathbf{n} = 0\).

(b) The perpendicular distance from the origin to the plane \(\Pi\) is given by \(\frac{|1|}{\sqrt{1^2 + (-3)^2 + (-2)^2}} = \frac{1}{\sqrt{14}}\).

(c) The angle between the planes is found using the dot product of their normals. The normal to OAB is \(\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\) and for \(\Pi\) is \(\begin{pmatrix} 1 \\ -3 \\ -2 \end{pmatrix}\). The cosine of the angle \(\alpha\) is \(\cos \alpha = \frac{\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -3 \\ -2 \end{pmatrix}}{\sqrt{3} \sqrt{14}} = \frac{6}{\sqrt{42}}\). The angle is \(22.2^\circ\).

(d) To find the common perpendicular, we first find the direction vector by solving the system of equations derived from the dot product conditions. The direction vector is \(\begin{pmatrix} 5 \\ -3 \\ 1 \end{pmatrix}\). The position vector is \(\begin{pmatrix} -0.1 \\ 0.7 \\ 0.3 \end{pmatrix}\). Thus, the equation is \(\mathbf{r} = \begin{pmatrix} -0.1 \\ 0.7 \\ 0.3 \end{pmatrix} + \lambda \begin{pmatrix} 5 \\ -3 \\ 1 \end{pmatrix}\).