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Let \((1+ax)^n = 1 + \binom{n}{1}(ax) + \binom{n}{2}(ax)^2 + \binom{n}{3}(ax)^3 + \dots\). Given that the coefficient of \(x\) is \(-16\), we have \(na = -16\) ... (1). Given that the coefficient of \(x^2\) is \(120\), we have \(\frac{n(n-1)}{2} a^2 = 120\) ... (2). From (1), \(a = -\frac{16}{n}\). Substituting this into (2), we get \(\frac{n(n-1)}{2} \left(-\frac{16}{n}\right)^2 = 120\), which simplifies to \(\frac{128(n-1)}{n} = 120\). Solving this yields \(128n - 128 = 120n \implies 8n = 128 \implies n = 16\). Substituting \(n = 16\) back into (1) gives \(a = -1\). The coefficient of \(x^3\) is given by \(\binom{n}{3} a^3 = \binom{16}{3} (-1)^3 = -560\).

1M for setting up \(na = -16\) and \(\frac{n(n-1)a^2}{2} = 120\); 1M for solving the rational equation in terms of \(n\); 1A for \(n=16\); 1A for \(a=-1\); 1M for attempting to evaluate \(\binom{n}{3}a^3\); 1.25A for finding the final answer \(-560\).

Let \(f(x) = \sqrt{5-2x}\). By first principles, the derivative is \(\lim_{h\to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h\to 0} \frac{\sqrt{5-2(x+h)} - \sqrt{5-2x}}{h}\). Multiplying the numerator and denominator by the conjugate \(\sqrt{5-2(x+h)} + \sqrt{5-2x}\), we get \(\lim_{h\to 0} \frac{(5-2x-2h) - (5-2x)}{h(\sqrt{5-2x-2h} + \sqrt{5-2x})} = \lim_{h\to 0} \frac{-2h}{h(\sqrt{5-2x-2h} + \sqrt{5-2x})}\). Canceling \(h\) yields \(\lim_{h\to 0} \frac{-2}{\sqrt{5-2x-2h} + \sqrt{5-2x}}\). Taking the limit as \(h \to 0\), we obtain \(\frac{-2}{2\sqrt{5-2x}} = -\frac{1}{\sqrt{5-2x}}\).

1M for using the limit definition of derivative; 1M for multiplying by the rational conjugate; 1M for expanding and simplifying the numerator to \(-2h\); 1.25M for canceling \(h\) properly; 2A for evaluating the limit correctly and complete proof steps with proper limit notation.

Let \(P(n)\) be the statement \(\sum_{r=1}^n \frac{1}{(2r-1)(2r+1)} = \frac{n}{2n+1}\). For \(n=1\), \(\text{LHS} = \frac{1}{1 \times 3} = \frac{1}{3}\) and \(\text{RHS} = \frac{1}{2(1)+1} = \frac{1}{3}\). Since \(\text{LHS} = \text{RHS}\), \(P(1)\) is true. Assume \(P(k)\) is true for some positive integer \(k\), i.e., \(\sum_{r=1}^k \frac{1}{(2r-1)(2r+1)} = \frac{k}{2k+1}\). For \(n=k+1\), \(\sum_{r=1}^{k+1} \frac{1}{(2r-1)(2r+1)} = \sum_{r=1}^k \frac{1}{(2r-1)(2r+1)} + \frac{1}{(2k+1)(2k+3)} = \frac{k}{2k+1} + \frac{1}{(2k+1)(2k+3)}\). Combining the fractions, we get \(\frac{k(2k+3) + 1}{(2k+1)(2k+3)} = \frac{2k^2 + 3k + 1}{(2k+1)(2k+3)} = \frac{(2k+1)(k+1)}{(2k+1)(2k+3)} = \frac{k+1}{2(k+1)+1}\). Hence, \(P(k+1)\) is true. By mathematical induction, \(P(n)\) is true for all positive integers \(n\).

1M for showing the case \(n=1\) is true; 1M for declaring the inductive hypothesis; 1M for adding the \((k+1)\)-th term; 2M for algebraic factorisation and simplification; 1.25A for concluding the induction proof.

(a) \(A^2 = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 9-4 & -12+4 \\ 3-1 & -4+1 \end{pmatrix} = \begin{pmatrix} 5 & -8 \\ 2 & -3 \end{pmatrix}\). (b) Let \(P(n)\) be the statement \(A^n = \begin{pmatrix} 2n+1 & -4n \\ n & 1-2n \end{pmatrix}\). For \(n=1\), \(\text{LHS} = A^1 = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\) and \(\text{RHS} = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\). Since \(\text{LHS} = \text{RHS}\), \(P(1)\) is true. Assume \(P(k)\) is true for some positive integer \(k\), i.e., \(A^k = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}\). For \(n=k+1\), \(A^{k+1} = A^k A = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 3(2k+1)-4k & -4(2k+1)+4k \\ 3k+1-2k & -4k-(1-2k) \end{pmatrix} = \begin{pmatrix} 2k+3 & -4k-4 \\ k+1 & -2k-1 \end{pmatrix} = \begin{pmatrix} 2(k+1)+1 & -4(k+1) \\ k+1 & 1-2(k+1) \end{pmatrix}\). Hence, \(P(k+1)\) is true. By mathematical induction, \(P(n)\) is true for all positive integers \(n\).

1.25A for finding \(A^2\); 1M for showing the case \(n=1\) is true; 1M for declaring the inductive hypothesis; 1M for calculating the product \(A^k A\); 1M for simplification of elements; 1A for concluding the induction proof.

Let \(h\text{ cm}\) be the depth of water, \(r\text{ cm}\) be the radius of the water surface, and \(V\text{ cm}^3\) be the volume of water in the vessel at time \(t\text{ s}\). By similar triangles, \(\frac{r}{h} = \frac{6}{12} = \frac{1}{2} \implies r = \frac{h}{2}\). The volume of water is \(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi}{12} h^3\). Differentiating with respect to \(t\), we have \(\frac{dV}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}\). Substituting \(\frac{dV}{dt} = 3\pi\) and \(h = 4\), we get \(3\pi = \frac{\pi}{4} (4)^2 \frac{dh}{dt} \implies 3\pi = 4\pi \frac{dh}{dt} \implies \frac{dh}{dt} = 0.75\text{ cm/s}\).

1.25M for expressing \(r\) in terms of \(h\) using similar triangles; 1M for expressing \(V\) as a function of \(h\) only; 2M for differentiating with respect to \(t\) using chain rule; 1M for substituting given values; 1A for finding \(0.75\text{ cm/s}\) (with units).

(a) Let \(I = \int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} dx\). Applying the substitution \(x = \pi - u\), we have \(dx = -du\). When \(x = 0\), \(u = \pi\); when \(x = \pi\), \(u = 0\). Then \(I = \int_{\pi}^0 \frac{(\pi - u) \sin(\pi - u)}{1 + \cos^2(\pi - u)} (-du) = \int_0^{\pi} \frac{(\pi - u) \sin u}{1 + \cos^2 u} du = \pi \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} dx - I\). This yields \(2I = \pi \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} dx \implies I = \frac{\pi}{2} \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} dx\). (b) To evaluate the integral, let \(t = \cos x\), which gives \(dt = -\sin x dx\). When \(x = 0\), \(t = 1\); when \(x = \pi\), \(t = -1\). Thus, \(\int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} dx = \int_1^{-1} \frac{-dt}{1 + t^2} = \int_{-1}^1 \frac{dt}{1 + t^2} = [\arctan t]_{-1}^1 = \frac{\pi}{4} - (-\frac{\pi}{4}) = \frac{\pi}{2}\). Hence, \(I = \frac{\pi}{2} \left(\frac{\pi}{2}\right) = \frac{\pi^2}{4}\).

1M for applying substitution \(x = \pi - u\); 1M for simplifying using trigonometric identities; 1A for completing proof of (a); 1M for substitution \(t = x\); 1.25M for integration and limits; 1A for final answer \(\frac{\pi^2}{4}\).

(a) The augmented matrix for system \((E)\) is \(\begin{pmatrix} 1 & 1 & 1 & 2 \\ 2 & 3 & k & 5 \\ 1 & 2 & 3 & 3 \end{pmatrix}\). Performing row operations \(R_2 \to R_2 - 2R_1\) and \(R_3 \to R_3 - R_1\) gives \(\begin{pmatrix} 1 & 1 & 1 & 2 \\ 0 & 1 & k-2 & 1 \\ 0 & 1 & 2 & 1 \end{pmatrix}\). Performing \(R_3 \to R_3 - R_2\) yields \(\begin{pmatrix} 1 & 1 & 1 & 2 \\ 0 & 1 & k-2 & 1 \\ 0 & 0 & 4-k & 0 \end{pmatrix}\). For the system to have infinitely many solutions, we must have \(4-k = 0 \implies k = 4\). (b) When \(k = 4\), the system simplifies to \(y + 2z = 1\) and \(x + y + z = 2\). Setting \(z = t\) where \(t\) is any real number, we get \(y = 1 - 2t\) and \(x = 2 - (1 - 2t) - t = 1 + t\). Thus, the solution is \(x = 1+t\), \(y = 1-2t\), \(z = t\) for any real number \(t\).

1M for setting up the augmented matrix; 1M for executing row operations correctly to upper triangular form; 1.25A for finding \(k = 4\); 1M for assigning parameter \(z = t\); 2A for solving \(x\) and \(y\) correctly in terms of \(t\).

The volume of the parallelepiped is given by \(| \vec{u} \cdot (\vec{v} \times \vec{w}) |\). First, calculate the scalar triple product \(\vec{u} \cdot (\vec{v} \times \vec{w}) = \det \begin{pmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ 3 & \lambda & 2 \end{pmatrix} = 1(-2 - 3\lambda) - 2(4 - 9) - 1(2\lambda + 3) = 5 - 5\lambda\). Since the volume is \(15\), we set \(|5 - 5\lambda| = 15 \implies |1 - \lambda| = 3\). This gives two cases: \(1 - \lambda = 3 \implies \lambda = -2\), or \(1 - \lambda = -3 \implies \lambda = 4\). Thus, the possible values of \(\lambda\) are \(-2\) and \(4\).

1M for relating the volume of the parallelepiped to the scalar triple product / determinant; 2M for expanding the determinant correctly to \(5-5\lambda\); 1M for establishing the absolute value equation; 2.25A for finding both correct answers (1.25A for one correct value, 1A for the other).

(a) The system \( (E) \) has a unique solution if and only if \( \Delta \neq 0 \).
\( \Delta = \begin{vmatrix} 1 & a & 1 \\ a & 1 & a-1 \\ 2 & 2a & a \end{vmatrix} = 1(a - 2a^2 + 2a) - a(a^2 - 2a + 2) + 2a^2 - 2 \)
\( \Delta = -a^3 + 2a^2 + a - 2 = -(a-1)(a+1)(a-2) \).
Setting \( \Delta = 0 \) gives \( a = 1, a = -1, \) or \( a = 2 \).
Thus, \( (E) \) has a unique solution for \( a \in \mathbb{R} \setminus \{-1, 1, 2\} \).

(b) (i) When \( a = 2 \), the augmented matrix is:
\( \begin{pmatrix} 1 & 2 & 1 & | & 1 \\ 2 & 1 & 1 & | & 2 \\ 2 & 4 & 2 & | & b \end{pmatrix} \)
By row operations: \( R_2 \to R_2 - 2R_1, R_3 \to R_3 - 2R_1 \):
\( \begin{pmatrix} 1 & 2 & 1 & | & 1 \\ 0 & -3 & -1 & | & 0 \\ 0 & 0 & 0 & | & b-2 \end{pmatrix} \).
For consistency, we require \( b - 2 = 0 \implies b = 2 \).

(ii) When \( b = 2 \), the system becomes:
\( \begin{cases} x + 2y + z = 1 \\ 3y + z = 0 \end{cases} \).
Let \( y = s \) where \( s \in \mathbb{R} \).
Then \( z = -3s \).
From the first equation, \( x = 1 - 2y - z = 1 - 2s - (-3s) = 1 + s \).
Thus, the solution is \( \{ (1+s, s, -3s) : s \in \mathbb{R} \} \).

(c) (i) When \( a = 1 \), the augmented matrix is:
\( \begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 1 & 1 & 0 & | & 1 \\ 2 & 2 & 1 & | & b \end{pmatrix} \)
By \( R_2 \to R_2 - R_1, R_3 \to R_3 - 2R_1 \):
\( \begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 0 & -1 & | & 0 \\ 0 & 0 & -1 & | & b-2 \end{pmatrix} \)
By \( R_3 \to R_3 - R_2 \):
\( \begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 0 & -1 & | & 0 \\ 0 & 0 & 0 & | & b-2 \end{pmatrix} \).
For consistency, we must have \( b - 2 = 0 \implies b = 2 \).

(ii) When \( b = 2 \), the system \( (F) \) is:
\( \begin{cases} x + y + z = 1 \\ x + y = 1 \\ 2x + 2y + z + c(x-y) = 2 \end{cases} \)
From the first two equations, \( z = 0 \) and \( x + y = 1 \implies y = 1 - x \).
Substituting into the third equation:
\( 2(1) + 0 + c(x - (1-x)) = 2 \implies c(2x - 1) = 0 \).
For the system to have infinitely many solutions, the relation must hold for any \( x \in \mathbb{R} \).
Thus, \( c = 0 \).

(a) Finding determinant: 1M. Setting determinant to 0: 1M. Correct answer: 1A.
(b)(i) Correct row operations: 1M. Correct value of b: 1A.
(b)(ii) Parametrizing variables: 1M. Correct solutions for x, y, and z: 2A.
(c)(i) Row operations for a=1: 1M. Correct value of b: 1A.
(c)(ii) Substituting relations: 1M. Analysis of the condition: 1M. Correct value of c: 1A.

(a) Let \( P(n) \) be the proposition: \( \sin \theta + \sin 3\theta + \dots + \sin(2n-1)\theta = \frac{\sin^2 n\theta}{\sin\theta} \).
When \( n = 1 \):
L.H.S. \( = \sin\theta \)
R.H.S. \( = \frac{\sin^2 \theta}{\sin\theta} = \sin\theta \).
Since L.H.S. = R.H.S., \( P(1) \) is true.

Assume \( P(k) \) is true for some positive integer \( k \), i.e.,
\( \sin \theta + \sin 3\theta + \dots + \sin(2k-1)\theta = \frac{\sin^2 k\theta}{\sin\theta} \).

For \( n = k+1 \):
L.H.S.
\( = \sin \theta + \sin 3\theta + \dots + \sin(2k-1)\theta + \sin(2k+1)\theta \)
\( = \frac{\sin^2 k\theta}{\sin\theta} + \sin(2k+1)\theta \) (by induction hypothesis)
\( = \frac{\sin^2 k\theta + \sin(2k+1)\theta \sin\theta}{\sin\theta} \)
Using product-to-sum identity: \( \sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)] \):
\( = \frac{\frac{1-\cos 2k\theta}{2} + \frac{1}{2}(\cos 2k\theta - \cos(2k+2)\theta)}{\sin\theta} \)
\( = \frac{1 - \cos(2k+2)\theta}{2\sin\theta} \)
\( = \frac{\sin^2(k+1)\theta}{\sin\theta} \) = R.H.S.
Thus, \( P(k+1) \) is true.
By the principle of mathematical induction, \( P(n) \) is true for all positive integers \( n \).

(b) (i) From (a), we have:
\( \sum_{r=1}^{n} \sin(2r-1)\theta = \frac{\sin^2 n\theta}{\sin\theta} \) and \( \sum_{r=1}^{n-1} \sin(2r-1)\theta = \frac{\sin^2 (n-1)\theta}{\sin\theta} \).
Subtracting these two equations gives:
\( \frac{\sin^2 n\theta}{\sin\theta} - \frac{\sin^2 (n-1)\theta}{\sin\theta} = \sin(2n-1)\theta \).

(ii) By putting \( n = 3 \) into (b)(i), we have:
\( \frac{\sin^2 3\theta - \sin^2 2\theta}{\sin\theta} = \sin 5\theta \).
Thus,
\( \int_{\pi/6}^{\pi/3} \frac{\sin^2 3\theta - \sin^2 2\theta}{\sin\theta} d\theta = \int_{\pi/6}^{\pi/3} \sin 5\theta d\theta \)
\( = \left[ -\frac{\cos 5\theta}{5} \right]_{\pi/6}^{\pi/3} \)
\( = -\frac{1}{5} \left( \cos\frac{5\pi}{3} - \cos\frac{5\pi}{6} \right) \)
\( = -\frac{1}{5} \left( \frac{1}{2} - \left(-\frac{\sqrt{3}}{2}\right) \right) \)
\( = -\frac{1+\sqrt{3}}{10} \).

(a) Proving the case for n=1: 1M. Stating induction hypothesis: 1M. Writing down inductive step L.H.S.: 1M. Simplifying using trigonometric identities: 1M. Reaching conclusion: 1A.
(b)(i) Identifying terms using summation representation: 1M. Proof completed: 1A.
(b)(ii) Applying (b)(i) to simplify the integrand: 1M. Performing correct integration: 1M. Evaluating the boundaries: 2M. Correct final answer: 1A.

(a) First, find the derivative using the quotient rule:
\( f'(x) = \frac{(2x-3)(x-1) - (x^2-3x+6)(1)}{(x-1)^2} = \frac{x^2 - 2x - 3}{(x-1)^2} = \frac{(x-3)(x+1)}{(x-1)^2} \).
Set \( f'(x) = 0 \implies x = 3 \) or \( x = -1 \).
We also compute the second derivative to test the turning points:
\( f(x) = x - 2 + \frac{4}{x-1} \implies f'(x) = 1 - \frac{4}{(x-1)^2} \implies f''(x) = \frac{8}{(x-1)^3} \).
At \( x = 3 \), \( f''(3) = 1 > 0 \), so \( (3, f(3)) = (3, 3) \) is a local minimum point.
At \( x = -1 \), \( f''(-1) = -1 < 0 \), so \( (-1, f(-1)) = (-1, -5) \) is a local maximum point.

(b) Since \( \lim_{x \to 1^+} f(x) = \infty \), the vertical asymptote is \( x = 1 \).
Using long division, \( f(x) = x - 2 + \frac{4}{x-1} \).
Since \( \lim_{x \to \pm\infty} (f(x) - (x-2)) = 0 \), the oblique asymptote is \( y = x - 2 \).

(c) For \( C \) to be concave upward, we need \( f''(x) > 0 \):
\( \frac{8}{(x-1)^3} > 0 \implies x > 1 \).
For \( C \) to be concave downward, we need \( f''(x) < 0 \):
\( \frac{8}{(x-1)^3} < 0 \implies x < 1 \).

(d) The sketch should clearly show:
1. The vertical asymptote \( x = 1 \) and oblique asymptote \( y = x - 2 \).
2. The turning points \( (-1, -5) \) and \( (3, 3) \).
3. Correct concavity (downward for \( x < 1 \) and upward for \( x > 1 \)).

(a) Finding f'(x): 1M. Setting f'(x)=0: 1M. Confirming types of turning points: 1M. Correct coordinates of local max and min: 1A.
(b) Correct vertical asymptote: 1A. Formulating oblique asymptote: 1M. Correct oblique asymptote: 1A.
(c) Correct range of concave upward: 1A. Correct range of concave downward: 1A.
(d) Sketching asymptotes: 1A. Sketching turning points: 1A. General correct curve shape: 1A.

(a) First, find the vectors:
\( \overrightarrow{AB} = (3-2)\mathbf{i} + (-1-1)\mathbf{j} + (2-(-1))\mathbf{k} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \).
\( \overrightarrow{AC} = (1-2)\mathbf{i} + (2-1)\mathbf{j} + (k-(-1))\mathbf{k} = -\mathbf{i} + \mathbf{j} + (k+1)\mathbf{k} \).
Now compute the cross product:
\( \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & 3 \\ -1 & 1 & k+1 \end{vmatrix} \)
\( = \mathbf{i} [-2(k+1) - 3] - \mathbf{j} [1(k+1) - (-3)] + \mathbf{k} [1(1) - 2] \)
\( = (-2k-5)\mathbf{i} - (k+4)\mathbf{j} - \mathbf{k} \).

(b) (i) Area of triangle \( ABC = \frac{1}{2} \| \overrightarrow{AB} \times \overrightarrow{AC} \| = \frac{3\sqrt{6}}{2} \implies \| \overrightarrow{AB} \times \overrightarrow{AC} \|^2 = 54 \).
\( (-2k-5)^2 + (k+4)^2 + (-1)^2 = 54 \)
\( (4k^2 + 20k + 25) + (k^2 + 8k + 16) + 1 = 54 \)
\( 5k^2 + 28k + 42 = 54 \implies 5k^2 + 28k - 12 = 0 \)
\( (5k-2)(k+6) = 0 \implies k = \frac{2}{5} \) or \( k = -6 \).

(ii) The integer value of \( k \) is \( -6 \).
Substituting \( k = -6 \) into \( \overrightarrow{AB} \times \overrightarrow{AC} \):
\( \mathbf{n} = (-2(-6)-5)\mathbf{i} - (-6+4)\mathbf{j} - \mathbf{k} = 7\mathbf{i} + 2\mathbf{j} - \mathbf{k} \).
This is normal to the plane \( \Pi \).
Using point \( A(2, 1, -1) \), the equation of the plane \( \Pi \) is:
\( 7(x-2) + 2(y-1) - 1(z+1) = 0 \implies 7x + 2y - z = 17 \).

(c) (i) First, find the vector \( \overrightarrow{AD} \):
\( \overrightarrow{AD} = (1-2)\mathbf{i} + (-1-1)\mathbf{j} + (4-(-1))\mathbf{k} = -\mathbf{i} - 2\mathbf{j} + 5\mathbf{k} \).
The volume \( V \) of the tetrahedron is given by:
\( V = \frac{1}{6} | (\overrightarrow{AB} \times \overrightarrow{AC}) \cdot \overrightarrow{AD} | \).
For \( k = -6 \):
\( (\overrightarrow{AB} \times \overrightarrow{AC}) \cdot \overrightarrow{AD} = (7\mathbf{i} + 2\mathbf{j} - \mathbf{k}) \cdot (-\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}) = -7 - 4 - 5 = -16 \).
Thus, \( V = \frac{|-16|}{6} = \frac{8}{3} \).

(ii) The shortest distance \( d \) from \( D(1, -1, 4) \) to the plane \( 7x + 2y - z - 17 = 0 \) is:
\( d = \frac{|7(1) + 2(-1) - 4 - 17|}{\sqrt{7^2 + 2^2 + (-1)^2}} = \frac{|-16|}{\sqrt{54}} = \frac{16}{3\sqrt{6}} = \frac{8\sqrt{6}}{9} \).
(Alternatively: \( d = \frac{|(\overrightarrow{AB} \times \overrightarrow{AC}) \cdot \overrightarrow{AD}|}{\|\overrightarrow{AB} \times \overrightarrow{AC}\|} = \frac{16}{3\sqrt{6}} = \frac{8\sqrt{6}}{9} \)).

(a) Finding vectors AB and AC: 1M. Correct determinants components: 1M. Correct cross product: 1A.
(b)(i) Setting up area equation: 1M. Simplifying to quadratic equation in k: 1M. Correct values of k: 2A (1A for each value).
(b)(ii) Finding normal vector: 1M. Correct plane equation: 1A.
(c)(i) Finding vector AD: 1M. Correct volume of tetrahedron: 1A.
(c)(ii) Applying distance formula: 1M. Correct distance in simplified surd form: 1A.