(a) Let line \(l_1\) pass through \(\mathbf{a} = (t,1,0)\) with direction \(\mathbf{b} = (-2,-1,0)\), and line \(l_2\) pass through \(\mathbf{c} = (0,1,t)\) with direction \(\mathbf{d} = (0,-2,1)\).

The shortest distance between two skew lines is \[ \text{distance} = \frac{\big|(\mathbf{b} \times \mathbf{d}) \cdot (\mathbf{c}-\mathbf{a})\big|} {\big|\mathbf{b} \times \mathbf{d}\big|}. \] Here \(\mathbf{b} \times \mathbf{d} = (-1,\,2,\,4)\), so \(\big|\mathbf{b} \times \mathbf{d}\big| = \sqrt{(-1)^2 + 2^2 + 4^2} = \sqrt{21}.\)

Also \(\mathbf{c} - \mathbf{a} = (0-t,\,1-1,\,t-0) = (-t,\,0,\,t)\), so \[ (\mathbf{b} \times \mathbf{d}) \cdot (\mathbf{c}-\mathbf{a}) = (-1,2,4)\cdot(-t,0,t) = t + 4t = 5t. \] Therefore \[ \text{distance} = \frac{|5t|}{\sqrt{21}}. \]

Given the shortest distance is \(\sqrt{21}\), \[ \frac{|5t|}{\sqrt{21}} = \sqrt{21} \;\Rightarrow\; |5t| = 21 \;\Rightarrow\; |t| = \frac{21}{5}. \] Since \(t\) is positive, \(t = \dfrac{21}{5}.\)

(b) The plane \(\Pi_1\) contains \(l_1\) and is parallel to \(l_2\), so it contains the direction vectors of both lines: \((-2,-1,0)\) and \((0,-2,1)\), and passes through \((t,1,0) = \left(\dfrac{21}{5},1,0\right)\).

Hence an equation of \(\Pi_1\) is \[ \mathbf{r} = \frac{21}{5}\,\mathbf{i} + \mathbf{j} + \lambda(-2\mathbf{i} - \mathbf{j}) + \mu(-2\mathbf{j} + \mathbf{k}), \] where \(\lambda,\mu \in \mathbb{R}.\)

(c) The direction vector of \(l_2\) is \(\mathbf{d} = (0,-2,1)\) and a normal to \(\Pi_2\colon 5x - 6y + 7z = 0\) is \(\mathbf{n} = (5,-6,7)\).

The angle \(\alpha\) between a line and a plane satisfies \[ \sin\alpha = \frac{|\,\mathbf{d}\cdot\mathbf{n}\,|}{|\mathbf{d}|\,|\mathbf{n}|}. \] Here \(\mathbf{d}\cdot\mathbf{n} = 0\cdot5 + (-2)(-6) + 1\cdot7 = 19,\) \(|\mathbf{d}| = \sqrt{0^2+(-2)^2+1^2} = \sqrt{5},\) \(|\mathbf{n}| = \sqrt{5^2+(-6)^2+7^2} = \sqrt{110}.\)

So \[ \sin\alpha = \frac{19}{\sqrt{5}\,\sqrt{110}} = \frac{19}{\sqrt{550}}, \quad \alpha \approx 54.1^\circ. \]

(d) The acute angle between the planes \(\Pi_1\) and \(\Pi_2\) is the angle between their normals.

A normal to \(\Pi_1\) is \(\mathbf{n}_1 = \mathbf{b}\times\mathbf{d} = (-1,2,4)\), and a normal to \(\Pi_2\) is \(\mathbf{n}_2 = (5,-6,7)\).

Then \[ \cos\phi = \frac{|\,\mathbf{n}_1\cdot\mathbf{n}_2\,|} {|\mathbf{n}_1|\,|\mathbf{n}_2|} = \frac{|(-1,2,4)\cdot(5,-6,7)|} {\sqrt{21}\,\sqrt{110}} = \frac{11}{\sqrt{21\cdot110}} = \frac{11}{\sqrt{2310}}, \] so \(\phi \approx 76.8^\circ.\)

(a) To find the vertical asymptote, set the denominator equal to zero: \(x + 1 = 0\), giving \(x = -1\).

For the oblique asymptote, perform polynomial long division on \(\frac{x^2 + x + 9}{x + 1}\):

\(y = x + \frac{9}{x + 1}\), so the oblique asymptote is \(y = x\).

(b) To find stationary points, differentiate \(y = \frac{x^2 + x + 9}{x + 1}\) and set \(\frac{dy}{dx} = 0\).

\(\frac{dy}{dx} = 1 - \frac{9}{(x+1)^2} = 0 \Rightarrow (x+1)^2 = 9\).

Solving gives \(x + 1 = 3\) or \(x + 1 = -3\), so \(x = 2\) or \(x = -4\).

Substitute back to find \(y\):

For \(x = 2\), \(y = \frac{2^2 + 2 + 9}{2 + 1} = 5\), giving point \((2, 5)\).

For \(x = -4\), \(y = \frac{(-4)^2 - 4 + 9}{-4 + 1} = -7\), giving point \((-4, -7)\).

(a) Start with the identity for the difference of tangents:

\(\tan(r+1) - \tan r = \frac{\sin(r+1)\cos r - \cos(r+1)\sin r}{\cos(r+1)\cos r}\)

This simplifies to:

\(\frac{\sin 1}{\cos(r+1)\cos r}\)

Thus, \(\tan(r+1) - \tan r = \frac{\sin 1}{\cos(r+1)\cos r}\).

(b) Using the method of differences, we have:

\(\sum_{r=1}^{n} u_r = \sum_{r=1}^{n} \frac{1}{\cos(r+1)\cos r}\)

Recognizing the telescoping nature:

\(\sum_{r=1}^{n} \left( \tan(r+1) - \tan r \right) = \tan(n+1) - \tan 1\)

Thus, \(\sum_{r=1}^{n} u_r = \frac{\tan(n+1) - \tan 1}{\sin 1}\).

(c) As \(n \to \infty\), \(\tan(n+1)\) oscillates, so the series \(u_1 + u_2 + u_3 + \ldots\) does not converge.

(a) The value of \(S_1\) is given by the sum of the roots: \(S_1 = \alpha + \beta + \gamma = 2\).

To find \(S_2\), use the relation \(S_2 = S_1^2 - 2(0) = 4\).

(b)(i) The recurrence relation is \(S_{n+3} = 2S_{n+2} - \frac{3}{2}S_n\).

(b)(ii) Using the recurrence relation, \(S_4 = 2S_3 - \frac{3}{2}S_1 = 2(2S_2 - \frac{3}{2}S_0) - \frac{3}{2}S_1 = 4\).

(c) Substitute \(x = 2 - y\) into the cubic equation: \(2(2-y)^3 - 4(2-y)^2 + 3 = 0\). Simplifying gives \(2y^3 - 8y^2 + 8y - 3 = 0\).

(d) The value of \(\frac{1}{\alpha + \beta} + \frac{1}{\beta + \gamma} + \frac{1}{\gamma + \alpha}\) is \(\frac{8}{3}\).

(a) Base case: For \(n = 1\),

\(5 \times 1^4 + 1^2 = \frac{1}{2} (2)^2 (2+1) = 6\), so \(H_1\) is true.

Inductive step: Assume \(\sum_{r=1}^{k} (5r^4 + r^2) = \frac{1}{2} k^2 (k+1)^2 (2k+1)\).

Consider \(\sum_{r=1}^{k+1} (5r^4 + r^2) = \frac{1}{2} k^2 (k+1)^2 (2k+1) + 5(k+1)^4 + (k+1)^2\).

\(= \frac{1}{2} (k+1)^2 (2k^3 + k^2 + 10(k+1)^2 + 2)\)

\(= \frac{1}{2} (k+1)^2 (k^2 + 2k + 1)(2k+3)\)

So \(H_{k+1}\) is true. By induction, \(H_n\) is true for all positive integers \(n\).

(b) Using the result from part (a):

\(5 \sum_{r=1}^{n} r^4 + \frac{1}{6} n(n+1)(2n+1) = \frac{1}{2} n^2 (n+1)^2 (2n+1)\)

\(5 \sum_{r=1}^{n} r^4 = \frac{1}{6} n(n+1)(2n+1)(3n(n+1)-1)\)

\(\sum_{r=1}^{n} r^4 = \frac{1}{30} n(n+1)(2n+1)(3n+3n-1)\)