(a) To find the equation of the plane ABC, we first find the direction vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\(\overrightarrow{AB} = (2\mathbf{i} + 4\mathbf{j} - \mathbf{k}) - (2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) = 2\mathbf{j} - 5\mathbf{k}\)

\(\overrightarrow{AC} = (-3\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}) - (2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) = -5\mathbf{i} - 5\mathbf{j}\)

The normal vector \(\mathbf{n}\) to the plane is given by the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\):

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & -5 \\ -5 & -5 & 0 \end{vmatrix} = (-5(2) - 5(-5))\mathbf{i} + (0(-5) - (-5)(-5))\mathbf{j} + (0(-5) - (-5)(2))\mathbf{k}\)

\(= -10\mathbf{i} + 25\mathbf{j} + 10\mathbf{k}\)

\(= -5\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}\)

Using point \(A(2, 2, 4)\), the equation of the plane is:

\(-5(x - 2) + 5(y - 2) + 2(z - 4) = 0\)

\(-5x + 5y + 2z = 8\)

(b) The perpendicular distance from point \(D(2, 1, 3)\) to the plane is given by:

\(\frac{|(-5)(2) + 5(1) + 2(3) - 8|}{\sqrt{(-5)^2 + 5^2 + 2^2}} = \frac{7}{\sqrt{54}} \approx 0.953\)

(c) The shortest distance between the lines AB and CD is found using the vector \(\overrightarrow{CD}\) and the cross product:

\(\overrightarrow{CD} = (2\mathbf{i} + \mathbf{j} + 3\mathbf{k}) - (2\mathbf{i} + 4\mathbf{j} - \mathbf{k}) = -3\mathbf{j} + 4\mathbf{k}\)

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & -5 \\ 5 & 4 & -1 \end{vmatrix} = (18)\mathbf{i} + (-25)\mathbf{j} + (-10)\mathbf{k}\)

The shortest distance is:

\(\frac{|\overrightarrow{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{35}{\sqrt{1049}} \approx 1.08\)

(a) The vertical asymptote occurs where the denominator is zero: \(x + 2 = 0 \Rightarrow x = -2\).

The horizontal asymptote is found by dividing the leading coefficients: \(y = \frac{x^2 + ax + 1}{x + 2} = x + a - 2\).

(b) Differentiate \(y = \frac{x^2 + ax + 1}{x + 2}\) using the quotient rule:

\(\frac{dy}{dx} = \frac{(x+2)(2x+a) - (x^2+ax+1)}{(x+2)^2} = \frac{x^2 + 4x + 2a - 1}{(x+2)^2}\).

Set \(\frac{dy}{dx} = 0\):

\(x^2 + 4x + 2a - 1 = 0\).

The discriminant \(16 - 4(2a-1) = 20 - 8a < 0\) implies no real roots, hence no stationary points.

(c) The curve intersects the \(y\)-axis at \(x = 0\):

\(y = \frac{0^2 + a(0) + 1}{0 + 2} = \frac{1}{2}\).

Asymptotes: vertical at \(x = -2\), oblique at \(y = x + a - 2\).

(d)(i) Sketch \(y = \left| \frac{x^2 + ax + 1}{x + 2} \right|\) with the same asymptotes.

(ii) Draw the line \(y = a\) on the sketch.

(iii) Solve \(\left| \frac{x^2 + ax + 1}{x + 2} \right| < a\) for given intervals:

\(\frac{x^2 + ax + 1}{x + 2} = a\) or \(\frac{x^2 + ax + 1}{x + 2} = -a\).

\(x^2 + 1 - 2a = 0\) or \(x^2 + 2ax + 1 = 0\).

\(-a - \sqrt{a^2 - 2a - 1} < x < -\sqrt{2a - 1}\) and \(-a + \sqrt{a^2 - 2a - 1} < x < \sqrt{2a - 1}\).

Given intervals imply \(a = 5\).

(a) The polar coordinates of the point of C furthest from the pole are \((\sqrt{\pi \arctan \pi}, 0) = (1.99, 0)\).

(b) Using the substitution \(u = \pi - \theta\), the area \(A\) is given by:

\(A = \frac{1}{2} \int_0^\pi (\pi - \theta) \arctan(\pi - \theta) \, d\theta\)

or

\(A = \frac{1}{2} \int_0^\pi u \arctan(u) \, du\)

Integrating by parts:

\(\frac{1}{2} \left[ \frac{1}{2} (u^2) \arctan(u) \right]_0^\pi - \frac{1}{2} \int_0^\pi \frac{u^2}{1 + u^2} \, du\)

\(\int_0^\pi \frac{u^2}{1 + u^2} \, du = \left[ u - \arctan(u) \right]_0^\pi\)

\(A = \frac{1}{4} \pi^2 \arctan \pi - \frac{1}{4} (\pi^2 + 1) \arctan \pi - \frac{1}{4} \pi = 2.65\)

(c) Let \(y = r \sin \theta\), then:

\(\frac{dy}{d\theta} = \frac{d}{d\theta} \left( \sqrt{(\pi - \theta) \arctan(\pi - \theta)} \sin \theta \right)\)

\(= \frac{(\pi - \theta) \arctan(\pi - \theta) \cos \theta}{\sqrt{(\pi - \theta) \arctan(\pi - \theta)}} + \arctan(\pi - \theta) \sin \theta\)

Setting \(\frac{dy}{d\theta} = 0\) gives:

\(2(\pi - \theta) \arctan(\pi - \theta) \cot \theta - \frac{\pi - \theta}{1 + (\pi - \theta)^2} - \arctan(\pi - \theta) = 0\)

For \(\theta = 1.2\),

\(2(\pi - 1.2) \arctan(\pi - 1.2) \cot 1.2 - \frac{\pi - 1.2}{1 + (\pi - 1.2)^2} - \arctan(\pi - 1.2) = 0.151\)

For \(\theta = 1.3\),

\(2(\pi - 1.3) \arctan(\pi - 1.3) \cot 1.3 - \frac{\pi - 1.3}{1 + (\pi - 1.3)^2} - \arctan(\pi - 1.3) = -0.395\)

There is a sign change, confirming a root exists between \(\theta = 1.2\) and \(\theta = 1.3\).

(a) To show that A is non-singular, we need to show that its determinant is non-zero for all real \(k\).

The determinant of \(A\) is calculated as follows:

\(\det(A) = k \begin{vmatrix} 5 & 2 \\ 3 & -k \end{vmatrix} - 1 \begin{vmatrix} 6 & 2 \\ -1 & -k \end{vmatrix} + 0 \begin{vmatrix} 6 & 5 \\ -1 & 3 \end{vmatrix}\)

\(= k(5(-k) - 6) - (6(-k) + 2)\)

\(= k(-5k - 6) - (-6k + 2)\)

\(= -5k^2 - 6k + 6k - 2\)

\(= -5k^2 - 2\)

Since \(-5k^2 - 2\) is always negative for all real \(k\), the determinant is never zero. Therefore, A is non-singular.

(b) Given \(A^{-1} = \begin{pmatrix} 3 & 0 & -1 \\ 1 & 0 & 0 \\ -\frac{23}{2} & \frac{1}{2} & 3 \end{pmatrix}\), we use the property \(AA^{-1} = I\).

From the first row of \(A\) and the first column of \(A^{-1}\), we have:

\(k \cdot 3 + 1 \cdot 1 + 0 \cdot -\frac{23}{2} = 1\)

\(3k + 1 = 1\)

\(3k = 0\)

\(k = 0\)

(a) Let \(y = \alpha^2 + 1\). Then \(\alpha^2 = y - 1\). Substitute into the original equation:

\((y - 1)^3 + 2(y - 1)^2 + 3(y - 1) + 1 = 0\)

Expanding and simplifying gives:

\(y^3 - y^2 + 4y - 5 = 0\)

(b) Using the result from part (a), we have:

\((\alpha^2 + 1)^2 + (\beta^2 + 1)^2 + (\gamma^2 + 1)^2 = 1 - 2(4) = -7\)

(c) Again using the result from part (a):

\((\alpha^2 + 1)^3 + (\beta^2 + 1)^3 + (\gamma^2 + 1)^3 = -7 - 4 + 5(3) = 4\)