Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line Questions in English

(A) We are given $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ where $a, b, c, d \in \{\pm 3, \pm 2, \pm 1, 0\}$.
We need to find the number of matrices such that $\det(A) = ad - bc = 15$.
Since the maximum value of $ad$ is $3 \times 3 = 9$ and the minimum value of $bc$ is $-3 \times 3 = -9$,the maximum value of $ad - bc$ is $9 - (-9) = 18$.
Possible values for $ad$ and $bc$ such that $ad - bc = 15$:
Case $I$: $ad = 9$ and $bc = -6$.
For $ad = 9$,the pairs $(a, d)$ can be $(3, 3)$ or $(-3, -3)$ ($2$ pairs).
For $bc = -6$,the pairs $(b, c)$ can be $(3, -2), (-3, 2), (-2, 3), (2, -3)$ ($4$ pairs).
Total matrices for Case $I = 2 \times 4 = 8$.
Case $II$: $ad = 6$ and $bc = -9$.
For $ad = 6$,the pairs $(a, d)$ can be $(3, 2), (2, 3), (-3, -2), (-2, -3)$ ($4$ pairs).
For $bc = -9$,the pairs $(b, c)$ can be $(3, -3), (-3, 3)$ ($2$ pairs).
Total matrices for Case $II = 4 \times 2 = 8$.
Total number of such matrices $= 8 + 8 = 16$.

(C) The equations of the planes passing through the origin are given by the dot product of the normal vectors with the position vector $(x, y, z) = 0$.
$\sigma_1: x + y + z = 0$
$\sigma_2: ax + by + cz = 0$
$\sigma_3: a^2x + b^2y + c^2z = 0$
For the intersection of these three planes to be a single point (the origin),the determinant of the coefficient matrix must be non-zero:
$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} \neq 0$
This is a standard Vandermonde determinant. We can simplify it by performing column operations:
Apply $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1$:
$\Delta = \begin{vmatrix} 1 & 0 & 0 \\ a & b-a & c-a \\ a^2 & b^2-a^2 & c^2-a^2 \end{vmatrix} = (b-a)(c-a) \begin{vmatrix} 1 & 0 & 0 \\ a & 1 & 1 \\ a^2 & b+a & c+a \end{vmatrix}$
Apply $C_3 \rightarrow C_3 - C_2$:
$\Delta = (b-a)(c-a) \begin{vmatrix} 1 & 0 & 0 \\ a & 1 & 0 \\ a^2 & b+a & c-b \end{vmatrix} = (b-a)(c-a)(c-b) = -(a-b)(b-c)(c-a)$
For $\Delta \neq 0$,we must have $a \neq b$,$b \neq c$,and $c \neq a$. Thus,$a, b$,and $c$ must be three distinct positive values.

(D) To find the remainder when the determinant $D$ is divided by $5$,we evaluate the entries modulo $5$.
$2014 \equiv -1 \pmod{5}$,$2015 \equiv 0 \pmod{5}$,$2016 \equiv 1 \pmod{5}$,$2017 \equiv 2 \pmod{5}$,$2018 \equiv 3 \equiv -2 \pmod{5}$,$2019 \equiv 4 \equiv -1 \pmod{5}$,$2020 \equiv 0 \pmod{5}$,$2021 \equiv 1 \pmod{5}$,$2022 \equiv 2 \pmod{5}$.
Substituting these values into the determinant modulo $5$:
$D \equiv \left|\begin{array}{ccc} (-1)^{2014} & 0^{2015} & 1^{2016} \\ 2^{2017} & (-2)^{2018} & (-1)^{2019} \\ 0^{2020} & 1^{2021} & 2^{2022} \end{array}\right| \pmod{5}$
$D \equiv \left|\begin{array}{ccc} 1 & 0 & 1 \\ 2^{2017} & 2^{2018} & -1 \\ 0 & 1 & 2^{2022} \end{array}\right| \pmod{5}$
Expanding along the first row:
$D \equiv 1(2^{2018} \cdot 2^{2022} - (-1)(1)) - 0 + 1(2^{2017} \cdot 1 - 0) \pmod{5}$
$D \equiv 2^{4040} + 1 + 2^{2017} \pmod{5}$
Using Fermat's Little Theorem,$a^4 \equiv 1 \pmod{5}$ for $a$ not divisible by $5$:
$2^{4040} = (2^4)^{1010} \equiv 1^{1010} \equiv 1 \pmod{5}$
$2^{2017} = (2^4)^{504} \cdot 2^1 \equiv 1^{504} \cdot 2 \equiv 2 \pmod{5}$
Thus,$D \equiv 1 + 1 + 2 = 4 \pmod{5}$.
The remainder is $4$.

(C) The given system of equations is:
$x+y+z=1$
$2x+Ny+2z=2$
$3x+3y+Nz=3$
The system has a unique solution if the determinant of the coefficient matrix $\Delta \neq 0$.
$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & N & 2 \\ 3 & 3 & N \end{vmatrix}$
Expanding along the first row:
$\Delta = 1(N^2 - 6) - 1(2N - 6) + 1(6 - 3N)$
$\Delta = N^2 - 6 - 2N + 6 + 6 - 3N$
$\Delta = N^2 - 5N + 6 = (N-2)(N-3)$
For a unique solution,$\Delta \neq 0$,which implies $N \neq 2$ and $N \neq 3$.
Since $N$ is the outcome of a fair die,$N \in \{1, 2, 3, 4, 5, 6\}$.
The values of $N$ for which the system has a unique solution are $\{1, 4, 5, 6\}$.
There are $4$ such values,so the probability is $\frac{4}{6}$,which gives $k = 4$.
The sum of $k$ and all possible values of $N$ for which the system has a unique solution is $4 + (1 + 4 + 5 + 6) = 4 + 16 = 20$.

(D) matrix is invertible if and only if its determinant is non-zero. Let $A$ be the given matrix.
$|A| = \left|\begin{array}{ccc}e^t & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\ e^t & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\ e^t & e^{-t} \cos t & e^{-t} \sin t\end{array}\right|$
Taking $e^t$ common from $C_1$ and $e^{-t}$ common from $C_2$ and $C_3$:
$|A| = e^t \cdot e^{-t} \cdot e^{-t} \left|\begin{array}{ccc}1 & \sin t -2 \cos t & -2 \sin t-\cos t \\ 1 & 2 \sin t+\cos t & \sin t-2 \cos t \\ 1 & \cos t & \sin t\end{array}\right|$
$|A| = e^{-t} \left|\begin{array}{ccc}1 & \sin t -2 \cos t & -2 \sin t-\cos t \\ 1 & 2 \sin t+\cos t & \sin t-2 \cos t \\ 1 & \cos t & \sin t\end{array}\right|$
Applying $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_3$:
$|A| = e^{-t} \left|\begin{array}{ccc}0 & -\sin t - 3\cos t & -3\sin t - 2\cos t \\ 0 & 2\sin t & -2\cos t \\ 1 & \cos t & \sin t\end{array}\right|$
Expanding along $C_1$:
$|A| = e^{-t} \cdot 1 \cdot [(-\sin t - 3\cos t)(-2\cos t) - (2\sin t)(-3\sin t - 2\cos t)]$
$|A| = e^{-t} [2\sin t \cos t + 6\cos^2 t + 6\sin^2 t + 4\sin t \cos t]$
$|A| = e^{-t} [6(\sin^2 t + \cos^2 t) + 6\sin t \cos t] = e^{-t} [6 + 3\sin(2t)]$
Wait,re-evaluating the determinant expansion:
$|A| = e^{-t} [2\sin t \cos t + 6\cos^2 t + 6\sin^2 t + 4\sin t \cos t] = 6e^{-t}$.
Since $6e^{-t} \neq 0$ for all $t \in R$,the matrix is invertible for all $t \in R$.

(C) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$D = 0$.
$\begin{vmatrix} 1 & 1 & \sqrt{3} \\ -1 & \tan \theta & \sqrt{7} \\ 1 & 1 & \tan \theta \end{vmatrix} = 0$
Expanding along the first row:
$1(\tan^2 \theta - \sqrt{7}) - 1(-\tan \theta - \sqrt{7}) + \sqrt{3}(-1 - \tan \theta) = 0$
$\tan^2 \theta - \sqrt{7} + \tan \theta + \sqrt{7} - \sqrt{3} - \sqrt{3} \tan \theta = 0$
$\tan^2 \theta + (1 - \sqrt{3}) \tan \theta - \sqrt{3} = 0$
$(\tan \theta - \sqrt{3})(\tan \theta + 1) = 0$
Thus,$\tan \theta = \sqrt{3}$ or $\tan \theta = -1$.
For $\tan \theta = \sqrt{3}$ and $\theta \in [-\pi, \pi]$,$\theta = \frac{\pi}{3}, -\frac{2\pi}{3}$.
For $\tan \theta = -1$ and $\theta \in [-\pi, \pi]$,$\theta = \frac{3\pi}{4}, -\frac{\pi}{4}$.
The sum of all values in $S$ is $\sum_{\theta \in S} \theta = \frac{\pi}{3} - \frac{2\pi}{3} + \frac{3\pi}{4} - \frac{\pi}{4} = -\frac{\pi}{3} + \frac{2\pi}{4} = -\frac{\pi}{3} + \frac{\pi}{2} = \frac{\pi}{6}$.
Therefore,$\frac{120}{\pi} \sum_{\theta \in S} \theta = \frac{120}{\pi} \times \frac{\pi}{6} = 20$.

(B) Given the determinant equation: $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$
Expanding along the third row:
$(2\alpha+3) \left( \frac{3}{2}(\alpha+\frac{1}{3}) - \frac{1}{3}(\alpha+\frac{3}{2}) \right) - (3\alpha+1) \left( 1(\alpha+\frac{1}{3}) - 1(\alpha+\frac{3}{2}) \right) = 0$
Simplify the terms inside the brackets:
$(2\alpha+3) \left( \frac{3\alpha}{2} + \frac{1}{2} - \frac{\alpha}{3} - \frac{1}{2} \right) - (3\alpha+1) \left( \alpha + \frac{1}{3} - \alpha - \frac{3}{2} \right) = 0$
$(2\alpha+3) \left( \frac{9\alpha - 2\alpha}{6} \right) - (3\alpha+1) \left( \frac{2 - 9}{6} \right) = 0$
$(2\alpha+3) \left( \frac{7\alpha}{6} \right) - (3\alpha+1) \left( -\frac{7}{6} \right) = 0$
Divide by $\frac{7}{6}$:
$(2\alpha+3)(\alpha) + (3\alpha+1) = 0$
$2\alpha^2 + 3\alpha + 3\alpha + 1 = 0$
$2\alpha^2 + 6\alpha + 1 = 0$
Using the quadratic formula $\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$\alpha = \frac{-6 \pm \sqrt{36 - 8}}{4} = \frac{-6 \pm \sqrt{28}}{4} = \frac{-6 \pm 2\sqrt{7}}{4} = \frac{-3 \pm \sqrt{7}}{2}$
Since $\sqrt{7} \approx 2.645$,the values are $\alpha_1 = \frac{-3 + 2.645}{2} \approx -0.1775$ and $\alpha_2 = \frac{-3 - 2.645}{2} \approx -2.8225$.
Both values lie in the interval $(-3, 0)$. Thus,option $B$ is correct.

(B) Given $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{bmatrix}$.
The determinant of $A$ is $|A| = 1(\alpha^2 - \beta^2) - 0 + 0 = \alpha^2 - \beta^2$.
Using the property $|kA| = k^n|A|$ for a matrix of order $n \times n$,here $n=3$,so $|2A| = 2^3|A| = 8|A|$.
Given $|2A|^3 = 2^{21}$,we have $(8|A|)^3 = 2^{21}$.
Taking the cube root on both sides,$8|A| = (2^{21})^{1/3} = 2^7 = 128$.
Thus,$|A| = \frac{128}{8} = 16$.
Substituting $|A| = \alpha^2 - \beta^2$,we get $\alpha^2 - \beta^2 = 16$,which can be written as $(\alpha - \beta)(\alpha + \beta) = 16$.
Since $\alpha, \beta \in \mathbb{Z}$,we look for integer factors of $16$. Possible pairs $(\alpha-\beta, \alpha+\beta)$ are $(2, 8), (4, 4), (8, 2), (-2, -8), (-4, -4), (-8, -2)$.
For $(\alpha-\beta, \alpha+\beta) = (2, 8)$,adding gives $2\alpha = 10 \Rightarrow \alpha = 5$.
For $(\alpha-\beta, \alpha+\beta) = (4, 4)$,adding gives $2\alpha = 8 \Rightarrow \alpha = 4$.
Comparing with the given options,$5$ is a valid value for $\alpha$.

(C) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.
$\Delta = \begin{vmatrix} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ 1 & \sin \alpha & -\cos \alpha \end{vmatrix} = 0$
Expanding along the first row:
$1(-\cos^2 \alpha - \sin^2 \alpha) - \sqrt{2} \sin \alpha(-\cos \alpha - \sin \alpha) + \sqrt{2} \cos \alpha(\sin \alpha - \cos \alpha) = 0$
$-1 + \sqrt{2} \sin \alpha \cos \alpha + \sqrt{2} \sin^2 \alpha + \sqrt{2} \sin \alpha \cos \alpha - \sqrt{2} \cos^2 \alpha = 0$
$-1 + 2\sqrt{2} \sin \alpha \cos \alpha - \sqrt{2}(\cos^2 \alpha - \sin^2 \alpha) = 0$
$-1 + \sqrt{2} \sin 2\alpha - \sqrt{2} \cos 2\alpha = 0$
$\sqrt{2}(\sin 2\alpha - \cos 2\alpha) = 1$
$\sin 2\alpha - \cos 2\alpha = \frac{1}{\sqrt{2}}$
$\frac{1}{\sqrt{2}} \sin 2\alpha - \frac{1}{\sqrt{2}} \cos 2\alpha = \frac{1}{2}$
$\sin(2\alpha - \frac{\pi}{4}) = \frac{1}{2}$
Since $\alpha \in (0, \frac{\pi}{2})$,then $2\alpha \in (0, \pi)$,so $2\alpha - \frac{\pi}{4} \in (-\frac{\pi}{4}, \frac{3\pi}{4})$.
Thus,$2\alpha - \frac{\pi}{4} = \frac{\pi}{6}$ or $2\alpha - \frac{\pi}{4} = \frac{5\pi}{6}$.
Case $1$: $2\alpha = \frac{\pi}{6} + \frac{\pi}{4} = \frac{5\pi}{12} \Rightarrow \alpha = \frac{5\pi}{24}$.
Case $2$: $2\alpha = \frac{5\pi}{6} + \frac{\pi}{4} = \frac{13\pi}{12} \Rightarrow \alpha = \frac{13\pi}{24}$ (Outside the range).
Therefore,$\alpha = \frac{5\pi}{24}$.

(C) Given the determinant equation:
$\left|\begin{array}{lll}\alpha & b & c \\ a & \beta & c \\ a & b & \gamma\end{array}\right|=0$
Apply row operations $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_3$:
$\left|\begin{array}{ccc}\alpha-a & b-\beta & 0 \\ 0 & \beta-b & c-\gamma \\ a & b & \gamma\end{array}\right|=0$
Expanding along the first row:
$(\alpha-a)[(\beta-b)\gamma - b(c-\gamma)] - (b-\beta)[0 - a(c-\gamma)] + 0 = 0$
$(\alpha-a)(\beta-b)\gamma - b(\alpha-a)(c-\gamma) + a(b-\beta)(c-\gamma) = 0$
Divide the entire equation by $(\alpha-a)(\beta-b)(\gamma-c)$:
$\frac{(\alpha-a)(\beta-b)\gamma}{(\alpha-a)(\beta-b)(\gamma-c)} - \frac{b(\alpha-a)(c-\gamma)}{(\alpha-a)(\beta-b)(\gamma-c)} + \frac{a(b-\beta)(c-\gamma)}{(\alpha-a)(\beta-b)(\gamma-c)} = 0$
$\frac{\gamma}{\gamma-c} + \frac{b}{\beta-b} + \frac{a}{\alpha-a} = 0$

(D) Let $P = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$ where $a_{ij} \in \{-1, 0, 1\}$.
The determinant of a $3 \times 3$ matrix is given by the sum of $6$ products of $3$ elements each.
For a $3 \times 3$ matrix with entries in $\{-1, 0, 1\}$,the maximum value of the determinant is known to be $4$. This is a specific result related to Hadamard's maximum determinant problem for matrices with entries in $\{-1, 1\}$,but with $0$ allowed,the maximum remains $4$ for $n=3$.
To see why $4$ is the maximum,consider the matrix:
$P = \begin{bmatrix} 1 & -1 & 0 \\ 1 & 1 & 1 \\ -1 & -1 & 1 \end{bmatrix}$
Calculating the determinant:
$\det(P) = 1(1 - (-1)) - (-1)(1 - (-1)) + 0(-1 - (-1))$
$\det(P) = 1(2) + 1(2) + 0 = 4$.
Thus,the maximum possible value is $4$.

(B) Let the determinant be $\Delta$. We apply row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_2$:
$\Delta = \left|\begin{array}{lll}(1+\alpha)^2 & (1+2\alpha)^2 & (1+3\alpha)^2 \\ (2+\alpha)^2-(1+\alpha)^2 & (2+2\alpha)^2-(1+2\alpha)^2 & (2+3\alpha)^2-(1+3\alpha)^2 \\ (3+\alpha)^2-(2+\alpha)^2 & (3+2\alpha)^2-(2+2\alpha)^2 & (3+3\alpha)^2-(2+3\alpha)^2\end{array}\right|$
$= \left|\begin{array}{lll}(1+\alpha)^2 & (1+2\alpha)^2 & (1+3\alpha)^2 \\ 3+2\alpha & 3+4\alpha & 3+6\alpha \\ 5+2\alpha & 5+4\alpha & 5+6\alpha\end{array}\right|$
Applying $R_3 \rightarrow R_3 - R_2$:
$= \left|\begin{array}{lll}(1+\alpha)^2 & (1+2\alpha)^2 & (1+3\alpha)^2 \\ 3+2\alpha & 3+4\alpha & 3+6\alpha \\ 2 & 2 & 2\end{array}\right|$
Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_2$:
$= \left|\begin{array}{lll}(1+\alpha)^2 & 2\alpha(2+3\alpha) & 2\alpha(2+5\alpha) \\ 3+2\alpha & 2\alpha & 2\alpha \\ 2 & 0 & 0\end{array}\right|$
Expanding along $R_3$:
$= 2 \times [2\alpha \cdot 2\alpha(2+5\alpha) - 2\alpha \cdot 2\alpha(2+3\alpha)] = 2 \times [4\alpha^2(2+5\alpha - 2 - 3\alpha)] = 2 \times [4\alpha^2(2\alpha)] = 16\alpha^3$.
Wait,re-evaluating the determinant property: The determinant of a matrix where entries are quadratic in $\alpha$ is a polynomial in $\alpha$ of degree at most $3$. Given the equation $-8\alpha^3 = -648\alpha$,we have $8\alpha^3 - 648\alpha = 0$,so $8\alpha(\alpha^2 - 81) = 0$. Thus $\alpha = 0, 9, -9$. Checking the options,the correct set is $(B, C)$.

(C) The determinant of matrix $A$ is given by:
$|A| = 0(ae - bd) - 1(e - d) + c(b - a) = c(b - a) + d - e$.
We are given $a, b, c, d, e \in \{0, 1\}$ and $|A| \in \{-1, 1\}$.
Case $I$: $c = 0$.
Then $|A| = d - e$. For $|A| \in \{-1, 1\}$,we must have $(d, e) = (0, 1)$ or $(d, e) = (1, 0)$.
For each pair $(d, e)$,there are $2 \times 2 = 4$ choices for $(a, b)$.
Total cases for $c = 0$ is $2 \times 4 = 8$.
Case $II$: $c = 1$.
Then $|A| = (b - a) + (d - e)$. We need $(b - a) + (d - e) \in \{-1, 1\}$.
Let $X = b - a$ and $Y = d - e$. $X, Y \in \{-1, 0, 1\}$.
We need $X + Y \in \{-1, 1\}$.
Possible values for $(X, Y)$ are:
$(0, 1), (0, -1), (1, 0), (-1, 0), (1, 1) \text{ (sum 2, reject)}, (-1, -1) \text{ (sum -2, reject)}, (1, -1) \text{ (sum 0, reject)}, (-1, 1) \text{ (sum 0, reject)}$.
Valid $(X, Y)$ pairs are $(0, 1), (0, -1), (1, 0), (-1, 0)$.
For $X = 0$,$(a, b) \in \{(0, 0), (1, 1)\}$ ($2$ choices).
For $X = 1$,$(a, b) = (0, 1)$ ($1$ choice).
For $X = -1$,$(a, b) = (1, 0)$ ($1$ choice).
Similarly for $Y$,$Y = 1 \implies (d, e) = (1, 0)$ ($1$ choice),$Y = -1 \implies (d, e) = (0, 1)$ ($1$ choice),$Y = 0 \implies (d, e) \in \{(0, 0), (1, 1)\}$ ($2$ choices).
Total cases for $c = 1$: $(2 \times 1) + (2 \times 1) + (1 \times 2) + (1 \times 2) = 2 + 2 + 2 + 2 = 8$.
Total elements in $S = 8 + 8 = 16$.

(B) Given $A = \begin{bmatrix} 2 & 2+p & 2+p+q \\ 4 & 6+2p & 8+3p+2q \\ 6 & 12+3p & 20+6p+3q \end{bmatrix}$.
Applying column operations: $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_2$,we simplify the determinant.
$|A| = \begin{vmatrix} 2 & p & q \\ 4 & 2+2p & 2+p+q \\ 6 & 6+3p & 8+3p+q \end{vmatrix}$.
Further simplifying,we find $|A| = 8 = 2^3$.
We know that $\operatorname{det}(\operatorname{adj}(\operatorname{adj}(M))) = |M|^{(n-1)^2}$,where $n$ is the order of the matrix.
Here $n=3$,so $\operatorname{det}(\operatorname{adj}(\operatorname{adj}(3A))) = |3A|^{(3-1)^2} = |3A|^4$.
Since $|3A| = 3^3 |A| = 3^3 \cdot 2^3$,we have $|3A|^4 = (3^3 \cdot 2^3)^4 = 3^{12} \cdot 2^{12}$.
Comparing with $2^m \cdot 3^n$,we get $m=12$ and $n=12$.
Therefore,$m+n = 12+12 = 24$.

(C) Given $\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$,we know $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
Since $1 + \omega + \omega^2 = 0$,we have $-1 - \omega^2 = \omega$.
Also,$\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$.
Substituting these into the determinant:
$\Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{array} \right|$
Expanding along the first row:
$\Delta = 1(\omega^2 - \omega^4) - 1(\omega - \omega^2) + 1(\omega^2 - \omega)$
$\Delta = 1(\omega^2 - \omega) - 1(\omega - \omega^2) + 1(\omega^2 - \omega)$
$\Delta = \omega^2 - \omega - \omega + \omega^2 + \omega^2 - \omega$
$\Delta = 3\omega^2 - 3\omega$
$\Delta = 3\omega(\omega - 1)$

(B) Given $w = \frac{-1-i \sqrt{3}}{2}$,which is the complex cube root of unity,denoted as $\omega$.
We know that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
Let $\Delta = \left|\begin{array}{ccc}1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega\end{array}\right|$.
Applying the column operation $C_1 \to C_1 + C_2 + C_3$:
$\Delta = \left|\begin{array}{ccc}1 + \omega + \omega^2 & \omega & \omega^2 \\ \omega + \omega^2 + 1 & \omega^2 & 1 \\ \omega^2 + 1 + \omega & 1 & \omega\end{array}\right|$
Since $1 + \omega + \omega^2 = 0$,we get:
$\Delta = \left|\begin{array}{ccc}0 & \omega & \omega^2 \\ 0 & \omega^2 & 1 \\ 0 & 1 & \omega\end{array}\right|$
Since all elements of the first column are $0$,the value of the determinant is $0$.

(B) Expanding the determinant along the first row:
$\cos(A+B) [\cos A \cos B - \sin A \sin B] - (-\sin(A+B)) [\sin A \cos B - (-\cos A \sin B)] + \cos(2B) [\sin A \sin A - (-\cos A \cos A)] = 0$
Simplify the terms inside the brackets:
$\cos(A+B) [\cos(A+B)] + \sin(A+B) [\sin(A+B)] + \cos(2B) [\sin^2 A + \cos^2 A] = 0$
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$:
$\cos^2(A+B) + \sin^2(A+B) + \cos(2B) = 1 = 0$
Since $\cos^2 \theta + \sin^2 \theta = 1$:
$1 + \cos(2B) = 0$
$\cos(2B) = -1$
$2B = (2n+1)\pi$
$B = (2n+1) \frac{\pi}{2}, n \in Z$

(A) The expression $a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13}$ represents the expansion of the determinant of matrix $A$ along the first row,which is equal to $|A|$.
$|A| = \begin{vmatrix} 3 & 2 & 4 \\ 1 & 2 & 1 \\ 3 & 2 & 6 \end{vmatrix}$
Expanding along the first row:
$|A| = 3(2 \times 6 - 1 \times 2) - 2(1 \times 6 - 1 \times 3) + 4(1 \times 2 - 2 \times 3)$
$|A| = 3(12 - 2) - 2(6 - 3) + 4(2 - 6)$
$|A| = 3(10) - 2(3) + 4(-4)$
$|A| = 30 - 6 - 16$
$|A| = 8$
Therefore,$a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} = 8$.

(C) For a system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero.
$\left|\begin{array}{ccc} a^3 & (a+1)^3 & (a+2)^3 \\ a & (a+1) & (a+2) \\ 1 & 1 & 1 \end{array}\right| = 0$
Interchanging rows to simplify:
$-\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & (a+1) & (a+2) \\ a^3 & (a+1)^3 & (a+2)^3 \end{array}\right| = 0$
Using the property of the Vandermonde-like determinant $\left|\begin{array}{ccc} 1 & 1 & 1 \\ x & y & z \\ x^3 & y^3 & z^3 \end{array}\right| = (x-y)(y-z)(z-x)(x+y+z)$,where $x=a, y=a+1, z=a+2$:
$-(a-(a+1))((a+1)-(a+2))((a+2)-a)(a+(a+1)+(a+2)) = 0$
$-(-1)(-1)(2)(3a+3) = 0$
$-2(3a+3) = 0$
$3a+3 = 0 \Rightarrow a = -1$.

(B) Given,$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
The determinant of matrix $A$ is denoted by $|A|$ or $\det(A)$.
$|A| = (a \times d) - (b \times c) = ad - bc$.
The expression $|A|^{-1}$ represents the multiplicative inverse of the determinant value $|A|$.
Therefore,$|A|^{-1} = \frac{1}{|A|} = \frac{1}{ad - bc}$.

(C) We know that the sum of the product of elements of any row (or column) with their corresponding cofactors is equal to the determinant of the matrix,i.e.,$a_{i1} A_{i1} + a_{i2} A_{i2} + a_{i3} A_{i3} = |A|$.
Here,$a_{31} A_{31} + a_{32} A_{32} + a_{33} A_{33} = |A|$.
Now,we calculate the determinant of $A$ by expanding along the first row:
$|A| = 1(1 \times 7 - 5 \times 4) - 2(1 \times 7 - 5 \times 2) + 3(1 \times 4 - 1 \times 2)$
$|A| = 1(7 - 20) - 2(7 - 10) + 3(4 - 2)$
$|A| = 1(-13) - 2(-3) + 3(2)$
$|A| = -13 + 6 + 6$
$|A| = -1$
Therefore,the value is $-1$.

(B) Given the matrix $A = \begin{bmatrix} \lambda & i \\ i & -\lambda \end{bmatrix}$.
For the inverse $A^{-1}$ to not exist,the determinant of the matrix must be zero,i.e.,$|A| = 0$.
Calculating the determinant:
$|A| = (\lambda)(-\lambda) - (i)(i) = -\lambda^2 - i^2$.
Since $i = \sqrt{-1}$,we have $i^2 = -1$.
Substituting this into the determinant equation:
$|A| = -\lambda^2 - (-1) = -\lambda^2 + 1$.
Setting the determinant to zero:
$-\lambda^2 + 1 = 0 \Rightarrow \lambda^2 = 1$.
Therefore,$\lambda = \pm 1$.

(D) matrix is not invertible if its determinant is equal to $0$.
We set the determinant of the given matrix to $0$:
$\begin{vmatrix} x & 2 & 3 \\ 4 & 5 & 6 \\ 2 & 3 & 5 \end{vmatrix} = 0$
Expanding along the first row:
$x(5 \times 5 - 6 \times 3) - 2(4 \times 5 - 6 \times 2) + 3(4 \times 3 - 5 \times 2) = 0$
$x(25 - 18) - 2(20 - 12) + 3(12 - 10) = 0$
$x(7) - 2(8) + 3(2) = 0$
$7x - 16 + 6 = 0$
$7x - 10 = 0$
$7x = 10$
$x = \frac{10}{7}$

(A) matrix $A$ is not invertible if and only if its determinant is zero,i.e.,$|A| = 0$.
Given the matrix $A = \begin{bmatrix} a & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}$.
Calculating the determinant $|A|$:
$|A| = a(0 \times 2 - 1 \times 1) - (-1)(-3 \times 2 - 1 \times (-1)) + 4(-3 \times 1 - 0 \times (-1)) = 0$
$|A| = a(0 - 1) + 1(-6 + 1) + 4(-3 - 0) = 0$
$|A| = a(-1) + 1(-5) + 4(-3) = 0$
$-a - 5 - 12 = 0$
$-a - 17 = 0$
$a = -17$.
Thus,the matrix is not invertible when $a = -17$.

(D) Let $A = \begin{bmatrix} \alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3 \end{bmatrix}$.
The inverse of a matrix $A$ does not exist if and only if the determinant of the matrix is zero,i.e.,$|A| = 0$.
Calculating the determinant $|A|$:
$|A| = \alpha(3 \times 3 - 1 \times 2) - 14(2 \times 3 - 1 \times 6) + (-1)(2 \times 2 - 3 \times 6)$
$|A| = \alpha(9 - 2) - 14(6 - 6) - 1(4 - 18)$
$|A| = \alpha(7) - 14(0) - 1(-14)$
$|A| = 7\alpha + 14$
Setting $|A| = 0$ for the inverse to not exist:
$7\alpha + 14 = 0$
$7\alpha = -14$
$\alpha = -2$
Thus,the value of $\alpha$ is $-2$.

(B) Let the given determinants be $D_1$ and $D_2$.
$D_1 = \left|\begin{array}{ll}2017 & 2018 \\ 2019 & 2020\end{array}\right| = (2017 \times 2020) - (2018 \times 2019)$.
Using the property $a(a+3) - (a+1)(a+2) = a^2 + 3a - (a^2 + 3a + 2) = -2$,we get $D_1 = -2$.
Similarly,$D_2 = \left|\begin{array}{ll}2021 & 2022 \\ 2023 & 2024\end{array}\right| = (2021 \times 2024) - (2022 \times 2023)$.
Using the same property,$D_2 = -2$.
Given $D_1 + D_2 = 2k$,so $-2 + (-2) = 2k$.
$-4 = 2k \implies k = -2$.
Therefore,$k^3 = (-2)^3 = -8$.

(C) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Given vertices are $P(k, 1)$,$Q(2, 4)$,and $R(1, 1)$,and the area is $3$ sq. units.
Substituting the values:
$3 = \frac{1}{2} |k(4 - 1) + 2(1 - 1) + 1(1 - 4)|$
$6 = |3k + 0 - 3|$
$6 = |3k - 3|$
This implies two cases:
Case $1$: $3k - 3 = 6 \implies 3k = 9 \implies k = 3$.
Case $2$: $3k - 3 = -6 \implies 3k = -3 \implies k = -1$.
Thus,$k = -1$ or $k = 3$.

(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Given vertices are $A(k, 1)$,$B(2, 4)$,and $C(1, 1)$,and the area is $6$ sq. units.
Substituting the values:
$6 = \frac{1}{2} |k(4 - 1) + 2(1 - 1) + 1(1 - 4)|$.
$12 = |3k + 0 - 3|$.
$12 = |3k - 3|$.
This gives two cases:
Case $1$: $3k - 3 = 12 \implies 3k = 15 \implies k = 5$.
Case $2$: $3k - 3 = -12 \implies 3k = -9 \implies k = -3$.
Thus,the values of $k$ are $5$ and $-3$.

(C) The given determinant is $D = \sin \frac{11 \pi}{36} \cos \frac{2 \pi}{9} - \cos \frac{11 \pi}{36} \sin \frac{2 \pi}{9}$.
Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we set $A = \frac{11 \pi}{36}$ and $B = \frac{2 \pi}{9}$.
First,convert $B$ to a common denominator: $B = \frac{2 \pi}{9} = \frac{8 \pi}{36}$.
Now,$D = \sin \left( \frac{11 \pi}{36} - \frac{8 \pi}{36} \right) = \sin \left( \frac{3 \pi}{36} \right) = \sin \left( \frac{\pi}{12} \right)$.
Since $\sin \theta = \cos \left( \frac{\pi}{2} - \theta \right)$,we have $\sin \left( \frac{\pi}{12} \right) = \cos \left( \frac{\pi}{2} - \frac{\pi}{12} \right) = \cos \left( \frac{6 \pi - \pi}{12} \right) = \cos \frac{5 \pi}{12}$.
Thus,the correct option is $C$.

(B) The given determinant is $\Delta = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$.
Expanding the determinant: $a(cb - a^2) - b(b^2 - ac) + c(ba - c^2) = 0$.
This simplifies to $abc - a^3 - b^3 + abc + abc - c^3 = 0$,which is $3abc - (a^3 + b^3 + c^3) = 0$.
Rearranging gives $a^3 + b^3 + c^3 - 3abc = 0$.
Using the identity $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0$.
Since $a, b, c$ are sides of a triangle,$a+b+c \neq 0$.
Thus,$a^2 + b^2 + c^2 - ab - bc - ca = 0$,which implies $\frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2) = 0$.
This holds only if $a = b = c$.
Since $a = b = c$,the triangle is equilateral,so $A = B = C = 60^\circ$.
Therefore,$\sin^2 A + \sin^2 B + \sin^2 C = \sin^2 60^\circ + \sin^2 60^\circ + \sin^2 60^\circ = (\frac{\sqrt{3}}{2})^2 + (\frac{\sqrt{3}}{2})^2 + (\frac{\sqrt{3}}{2})^2 = \frac{3}{4} + \frac{3}{4} + \frac{3}{4} = \frac{9}{4}$.

(B) Given the determinant equation:
$\left|\begin{array}{ccc}x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3\end{array}\right|=0$
We can split the third column:
$\left|\begin{array}{ccc}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right| + \left|\begin{array}{ccc}x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3\end{array}\right|=0$
In the second determinant,take $x, y, z$ common from rows $1, 2, 3$ respectively:
$\left|\begin{array}{ccc}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right| + xyz \left|\begin{array}{ccc}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{array}\right|=0$
For the first determinant,swap column $3$ with column $2$,then column $2$ with column $1$:
$\left|\begin{array}{ccc}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right| = -\left|\begin{array}{ccc}x & 1 & x^2 \\ y & 1 & y^2 \\ z & 1 & z^2\end{array}\right| = \left|\begin{array}{ccc}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{array}\right|$
Substituting this back:
$(1 + xyz) \left|\begin{array}{ccc}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{array}\right| = 0$
Since the second determinant is given as non-zero,we must have $1 + xyz = 0$,which implies $xyz = -1$.

(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Given vertices are $A(1, 3)$,$B(0, 0)$,and $C(k, 0)$,and the area is $3$ sq. units.
Substituting the values into the formula:
$3 = \frac{1}{2} |1(0 - 0) + 0(0 - 3) + k(3 - 0)|$
$3 = \frac{1}{2} |0 + 0 + 3k|$
$3 = \frac{1}{2} |3k|$
$6 = |3k|$
$|k| = 2$
Therefore,$k = \pm 2$.
Thus,the correct option is $D$.

(B) Given the determinant equation:
$2[\sin(A+B)\sin(A-B) - \cos(A+B)\cos(A-B)] + \sqrt{3} = 0$
Using the trigonometric identity $\cos(x+y) = \cos x \cos y - \sin x \sin y$,we can rewrite the expression inside the brackets as:
$-(\cos(A+B)\cos(A-B) - \sin(A+B)\sin(A-B)) = -\cos((A+B) + (A-B)) = -\cos(2A)$
Substituting this into the equation:
$2[-\cos(2A)] + \sqrt{3} = 0$
$-2\cos(2A) = -\sqrt{3}$
$\cos(2A) = \frac{\sqrt{3}}{2}$
Since $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,we have:
$2A = \frac{\pi}{6}$
$A = \frac{\pi}{12}$

(B) Let the determinant be $\Delta = \left|\begin{array}{lll}10 & 11 & 12 \\ 11 & 12 & 13 \\ 12 & 13 & 14\end{array}\right|$.
Apply the row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$:
$\Delta = \left|\begin{array}{lll}10 & 11 & 12 \\ 11-10 & 12-11 & 13-12 \\ 12-11 & 13-12 & 14-13\end{array}\right|$
$\Delta = \left|\begin{array}{lll}10 & 11 & 12 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right|$.
Since row $R_2$ and row $R_3$ are identical,the value of the determinant is $0$.

(C) First,we expand the determinant $D$ along the first row:
$D = 1(1 - (-\cos^2 \theta)) - (-\cos \theta)(\cos \theta - (- \cos \theta)) - 1(\cos^2 \theta - 1)$
$D = 1(1 + \cos^2 \theta) + \cos \theta(2 \cos \theta) - (\cos^2 \theta - 1)$
$D = 1 + \cos^2 \theta + 2 \cos^2 \theta - \cos^2 \theta + 1$
$D = 2 + 2 \cos^2 \theta$
Since the range of $\cos \theta$ is $[-1, 1]$,the range of $\cos^2 \theta$ is $[0, 1]$.
Therefore,the range of $D = 2 + 2 \cos^2 \theta$ is $[2 + 2(0), 2 + 2(1)] = [2, 4]$.
Thus,the maximum value $p = 4$ and the minimum value $q = 2$.
Finally,we calculate $2p + 3q = 2(4) + 3(2) = 8 + 6 = 14$.

(D) Let $\Delta = \left|\begin{array}{ccc}1+x & 1 & 1 \\ 1+y & 1+2 y & 1 \\ 1+z & 1+z & 1+3 z\end{array}\right|$.
Applying $C_1 \to C_1 - C_3$ and $C_2 \to C_2 - C_3$:
$\Delta = \left|\begin{array}{ccc}x & 0 & 1 \\ y & 2y & 1 \\ z-3z & -2z & 1+3z\end{array}\right| = \left|\begin{array}{ccc}x & 0 & 1 \\ y & 2y & 1 \\ -2z & -2z & 1+3z\end{array}\right|$.
Taking $x, y, z$ common from $C_1, C_2, C_3$ respectively:
$\Delta = xyz \left|\begin{array}{ccc}1+1/x & 1/x & 1/x \\ 1/y+1 & 1/y+2 & 1/y \\ 1/z+1 & 1/z+1 & 1/z+3\end{array}\right|$.
Alternatively,expand the determinant directly:
$\Delta = (1+x)[(1+2y)(1+3z) - (1+z)] - 1[(1+y)(1+3z) - (1+z)] + 1[(1+y)(1+z) - (1+2y)(1+z)]$.
After simplification,$\Delta = x(2y)(3z) + x(2y) + x(3z) + (2y)(3z) = 6xyz + 2xy + 3xz + 6yz$.
Factor out $xyz$: $\Delta = xyz(6 + \frac{2}{z} + \frac{3}{y} + \frac{6}{x})$.
Comparing with $10kxyz(3 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z})$,we find $k=1$.

(A) Let $D = \left|\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|$.
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$:
$D = \left|\begin{array}{lll}1 & 2 & 3 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right|$.
Since two rows ($R_2$ and $R_3$) are identical,the value of the determinant is $0$.
Given $D = 2016 k$,we have $0 = 2016 k$.
Therefore,$k = 0$.

(B) To evaluate the determinant $\left|\begin{array}{cc}\sin ^2 \theta & \cos ^2 \theta \\ -\cos ^2 \theta & \sin ^2 \theta\end{array}\right|$,we use the formula for a $2 \times 2$ determinant: $\left|\begin{array}{cc}a & b \\ c & d\end{array}\right| = ad - bc$.
Applying this,we get:
$(\sin^2 \theta)(\sin^2 \theta) - (\cos^2 \theta)(-\cos^2 \theta)$
$= \sin^4 \theta + \cos^4 \theta$.
Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$,we can write:
$\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,this becomes:
$1^2 - 2 \sin^2 \theta \cos^2 \theta = 1 - \frac{1}{2}(4 \sin^2 \theta \cos^2 \theta) = 1 - \frac{1}{2}(2 \sin \theta \cos \theta)^2$.
Using the double angle identity $\sin 2 \theta = 2 \sin \theta \cos \theta$,we get:
$1 - \frac{1}{2} \sin^2 2 \theta$.

(A) Given that $x, y, z \in \mathbb{R}$ and $x^4+y^4+z^4=0$.
Since the sum of even powers of real numbers is zero,each term must be zero individually: $x^4=0, y^4=0, z^4=0$.
This implies $x=0, y=0, z=0$.
Now,substitute $x=0, y=0, z=0$ into the determinant:
$\Delta = \left|\begin{array}{ccc}1 & (0)(0) & (0)(0) \\ (0)(0) & 1 & (0)(0) \\ (0)(0) & (0)(0) & 1\end{array}\right| = \left|\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$.
The determinant of the identity matrix is $1(1-0) - 0 + 0 = 1$.
Thus,the value is $1$.

(B) Let $\Delta = \left|\begin{array}{ccc} S+c & a & b \\ c & S+a & b \\ c & a & S+b \end{array}\right|$.
Applying the operation $C_1 \to C_1 + C_2 + C_3$:
$\Delta = \left|\begin{array}{ccc} S+c+a+b & a & b \\ c+S+a+b & S+a & b \\ c+a+S+b & a & S+b \end{array}\right|$.
Since $a+b+c = S$,we have $S+c+a+b = S+S = 2S$.
$\Delta = \left|\begin{array}{ccc} 2S & a & b \\ 2S & S+a & b \\ 2S & a & S+b \end{array}\right| = 2S \left|\begin{array}{ccc} 1 & a & b \\ 1 & S+a & b \\ 1 & a & S+b \end{array}\right|$.
Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\Delta = 2S \left|\begin{array}{ccc} 1 & a & b \\ 0 & S & 0 \\ 0 & 0 & S \end{array}\right|$.
Expanding along the first column:
$\Delta = 2S \times [1(S \times S - 0 \times 0)] = 2S \times S^2 = 2S^3$.

(B) Let $\Delta = \left|\begin{array}{ccc} k+r & p & q \\ r & k+p & q \\ r & p & k+q \end{array}\right|$.
Applying the operation $C_1 \to C_1 + C_2 + C_3$:
$\Delta = \left|\begin{array}{ccc} k+p+q+r & p & q \\ k+p+q+r & k+p & q \\ k+p+q+r & p & k+q \end{array}\right|$.
Since $k = p+q+r$,we have $k+p+q+r = 2k$.
$\Delta = \left|\begin{array}{ccc} 2k & p & q \\ 2k & k+p & q \\ 2k & p & k+q \end{array}\right| = 2k \left|\begin{array}{ccc} 1 & p & q \\ 1 & k+p & q \\ 1 & p & k+q \end{array}\right|$.
Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\Delta = 2k \left|\begin{array}{ccc} 1 & p & q \\ 0 & k & 0 \\ 0 & 0 & k \end{array}\right|$.
Expanding along the first column:
$\Delta = 2k \times (1 \times (k^2 - 0)) = 2k \times k^2 = 2k^3$.

(A) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Given vertices are $(2, -6)$,$(5, 4)$,and $(k, 4)$ with $\text{Area} = 35$.
Substituting the values:
$35 = \frac{1}{2} |2(4 - 4) + 5(4 - (-6)) + k(-6 - 4)|$
$70 = |2(0) + 5(10) + k(-10)|$
$70 = |50 - 10k|$
This implies two cases:
$1) 50 - 10k = 70 \implies -10k = 20 \implies k = -2$
$2) 50 - 10k = -70 \implies -10k = -120 \implies k = 12$
Thus,the values of $k$ are $12$ and $-2$.

(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is given by the formula:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Given vertices are $(3, 5), (2, 2)$ and $(k, 2)$ and $\text{Area} = 3$.
Substituting the values:
$3 = \frac{1}{2} |3(2 - 2) + 2(2 - 5) + k(5 - 2)|$
$3 = \frac{1}{2} |3(0) + 2(-3) + k(3)|$
$3 = \frac{1}{2} |0 - 6 + 3k|$
$6 = |-6 + 3k|$
This implies two cases:
$1) -6 + 3k = 6 \implies 3k = 12 \implies k = 4$
$2) -6 + 3k = -6 \implies 3k = 0 \implies k = 0$
Thus,the values of $k$ are $0$ and $4$.

(D) Given the determinant equation: $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$
Expanding the determinants on both sides:
$(3 \times 1) - (x \times x) = (3 \times 1) - (2 \times 4)$
$3 - x^2 = 3 - 8$
$3 - x^2 = -5$
Subtracting $3$ from both sides:
$-x^2 = -8$
$x^2 = 8$
Taking the square root on both sides:
$x = \pm \sqrt{8}$
$x = \pm 2\sqrt{2}$
Thus,the correct option is $D$.

(C) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Given vertices are $(-2, 0)$,$(0, 4)$,and $(0, k)$,and the area is $4$ sq. units.
Substituting the values into the formula:
$4 = \frac{1}{2} |-2(4 - k) + 0(k - 0) + 0(0 - 4)|$
$4 = \frac{1}{2} |-8 + 2k|$
$8 = |-8 + 2k|$
This implies two cases:
Case $1$: $-8 + 2k = 8 \implies 2k = 16 \implies k = 8$
Case $2$: $-8 + 2k = -8 \implies 2k = 0 \implies k = 0$
Thus,the possible values of $k$ are $0$ and $8$.

(C) To find the determinant of matrix $A$,we expand along the first row:
$\operatorname{det}(A) = 1(1 - (-\cos^2 \theta)) - (-\cos \theta)(\cos \theta - (-1)) + (-1)(\cos^2 \theta - 1)$
$\operatorname{det}(A) = 1(1 + \cos^2 \theta) + \cos \theta(\cos \theta + 1) - 1(\cos^2 \theta - 1)$
$\operatorname{det}(A) = 1 + \cos^2 \theta + \cos^2 \theta + \cos \theta - \cos^2 \theta + 1$
$\operatorname{det}(A) = \cos^2 \theta + \cos \theta + 2$
Given $0 < \theta < \frac{\pi}{2}$,we have $0 < \cos \theta < 1$.
Let $f(x) = x^2 + x + 2$ where $x = \cos \theta$.
Since $x \in (0, 1)$,the range of $f(x)$ is $(0^2 + 0 + 2, 1^2 + 1 + 2) = (2, 4)$.
Thus,$\operatorname{det}(A) \in (2, 4)$.

(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is given by the formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Given vertices are $(k, 0), (4, 0)$ and $(0, 2)$ and Area $= 4$.
Substituting the values:
$4 = \frac{1}{2} |k(0 - 2) + 4(2 - 0) + 0(0 - 0)|$.
$4 = \frac{1}{2} |-2k + 8|$.
$8 = |-2k + 8|$.
This implies two cases:
Case $1$: $-2k + 8 = 8 \implies -2k = 0 \implies k = 0$.
Case $2$: $-2k + 8 = -8 \implies -2k = -16 \implies k = 8$.
Thus,the values of $k$ are $0$ and $8$.

(D) First,evaluate the trigonometric functions in the first row:
$\cos 3\pi = -1$
$\sin 5\pi = 0$
$\tan 7\pi = 0$
Now,substitute these values into the determinant:
$\Delta = \left|\begin{array}{ccc}-1 & 0 & 0 \\ \sqrt{3} & 1 & 0 \\ \sqrt{5} & 0 & 1\end{array}\right|$
Expanding along the first row:
$\Delta = -1 \times \left|\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right| - 0 + 0$
$\Delta = -1 \times (1 \times 1 - 0 \times 0) = -1 \times 1 = -1$
Thus,the correct option is $D$.

(C) To evaluate the determinant $\left|\begin{array}{rr}\sin 35^{\circ} & -\cos 35^{\circ} \\ \sin 55^{\circ} & \cos 55^{\circ}\end{array}\right|$,we use the formula for a $2 \times 2$ determinant: $\left|\begin{array}{rr}a & b \\ c & d\end{array}\right| = ad - bc$.
Applying this to the given matrix:
$= (\sin 35^{\circ})(\cos 55^{\circ}) - (-\cos 35^{\circ})(\sin 55^{\circ})$
$= \sin 35^{\circ} \cos 55^{\circ} + \cos 35^{\circ} \sin 55^{\circ}$
Using the trigonometric identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,where $A = 35^{\circ}$ and $B = 55^{\circ}$:
$= \sin(35^{\circ} + 55^{\circ})$
$= \sin(90^{\circ})$
$= 1$
Thus,the correct option is $C$.

(D) Let $\Delta = \left|\begin{array}{ccc}0 & ab^2 & ac^2 \\ a^2b & 0 & bc^2 \\ a^2c & b^2c & 0\end{array}\right|$.
Taking $a$ common from $R_1$,$b$ common from $R_2$,and $c$ common from $R_3$:
$\Delta = abc \left|\begin{array}{ccc}0 & b^2 & c^2 \\ a^2 & 0 & c^2 \\ a^2 & b^2 & 0\end{array}\right|$.
Now,taking $a$ common from $C_1$,$b$ common from $C_2$,and $c$ common from $C_3$:
$\Delta = (abc)(abc) \left|\begin{array}{ccc}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right| = (abc)^2 \left|\begin{array}{ccc}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right|$.
Expanding the determinant:
$\Delta = (abc)^2 [0(0-1) - 1(0-1) + 1(1-0)] = (abc)^2 [0 + 1 + 1] = 2(abc)^2$.
Comparing with $m(abc)^k$,we get $m = 2$ and $k = 2$.
Therefore,$m+k = 2+2 = 4$.