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

(B) Given that $A$ and $B$ are square matrices of order $n = 3$.
We know the property of determinants that $|AB| = |A| \times |B|$.
So,$|AB| = (-1) \times (3) = -3$.
We also know the property $|kA| = k^n |A|$,where $n$ is the order of the matrix.
Here,$|3AB| = 3^3 |AB|$.
Substituting the value of $|AB|$,we get $|3AB| = 27 \times (-3) = -81$.

(A) To find the determinant of matrix $A$,we expand along the first row:
$|A| = \begin{vmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{vmatrix}$
$|A| = 1 \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} - 0 \begin{vmatrix} 2 & 0 \\ 3 & 1 \end{vmatrix} + 1 \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix}$
$|A| = 1(1 \times 1 - 0 \times 2) - 0 + 1(2 \times 2 - 1 \times 3)$
$|A| = 1(1 - 0) + 1(4 - 3)$
$|A| = 1(1) + 1(1) = 1 + 1 = 2$
Thus,$\det A = 2$.

(A) Given that $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$.
The determinant of matrix $A$ is $|A| = \alpha^2 - (2 \times 2) = \alpha^2 - 4$.
We are given that $|A^3| = 125$.
Using the property of determinants $|A^n| = |A|^n$,we have $|A|^3 = 125$.
Since $125 = 5^3$,we get $|A|^3 = 5^3$,which implies $|A| = 5$.
Substituting the value of $|A|$,we get $\alpha^2 - 4 = 5$.
$\alpha^2 = 9$.
Therefore,$\alpha = \pm 3$.

(A) Given that $A$ and $B$ are $n \times n$ matrices such that $AB = O$,where $O$ is the zero matrix.
Taking the determinant on both sides,we get $Det(AB) = Det(O)$.
Using the property $Det(AB) = Det(A) \times Det(B)$,we have $Det(A) \times Det(B) = 0$.
For the product of two numbers to be zero,at least one of them must be zero.
Therefore,$Det(A) = 0$ or $Det(B) = 0$.

(C) The determinant is given by $\Delta = \left| \begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array} \right|$.
Expanding along the first row:
$\Delta = a(cb - a^2) - b(b^2 - ca) + c(ab - c^2)$
$= abc - a^3 - b^3 + abc + abc - c^3$
$= 3abc - a^3 - b^3 - c^3$
$= -(a^3 + b^3 + c^3 - 3abc)$.
We know the algebraic identity: $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.
Substituting this into our expression:
$\Delta = -1 \times (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.
Comparing this with the given form $k(a + b + c)(a^2 + b^2 + c^2 - bc - ca - ab)$,we get $k = -1$.

(A) Given the determinant equation: $\left| \begin{array}{ccc} x + \alpha & \beta & \gamma \\ \gamma & x + \beta & \alpha \\ \alpha & \beta & x + \gamma \end{array} \right| = 0$.
Applying the column operation $C_1 \to C_1 + C_2 + C_3$:
$\left| \begin{array}{ccc} x + \alpha + \beta + \gamma & \beta & \gamma \\ x + \alpha + \beta + \gamma & x + \beta & \alpha \\ x + \alpha + \beta + \gamma & \beta & x + \gamma \end{array} \right| = 0$.
Taking $(x + \alpha + \beta + \gamma)$ common from $C_1$:
$(x + \alpha + \beta + \gamma) \left| \begin{array}{ccc} 1 & \beta & \gamma \\ 1 & x + \beta & \alpha \\ 1 & \beta & x + \gamma \end{array} \right| = 0$.
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$(x + \alpha + \beta + \gamma) \left| \begin{array}{ccc} 1 & \beta & \gamma \\ 0 & x & \alpha - \gamma \\ 0 & 0 & x \end{array} \right| = 0$.
Expanding the determinant along $C_1$:
$(x + \alpha + \beta + \gamma) [1(x^2 - 0)] = 0$.
$x^2(x + \alpha + \beta + \gamma) = 0$.
Therefore,the values of $x$ are $x = 0$ and $x = -(\alpha + \beta + \gamma)$.

(D) Given the determinant equation: $\left| \begin{array}{ccc} x & 3 & 7 \\ 2 & x & -2 \\ 7 & 8 & x \end{array} \right| = 0$.
Expanding along the first row:
$x(x^2 - (-16)) - 3(2x - (-14)) + 7(16 - 7x) = 0$
$x(x^2 + 16) - 3(2x + 14) + 7(16 - 7x) = 0$
$x^3 + 16x - 6x - 42 + 112 - 49x = 0$
$x^3 - 39x + 70 = 0$.
Since $x = 5$ is a root,$(x - 5)$ must be a factor.
Performing polynomial division or synthetic division on $x^3 - 39x + 70 = 0$:
$x^3 - 5x^2 + 5x^2 - 25x - 14x + 70 = 0$
$x^2(x - 5) + 5x(x - 5) - 14(x - 5) = 0$
$(x - 5)(x^2 + 5x - 14) = 0$
$(x - 5)(x + 7)(x - 2) = 0$.
Thus,the roots are $x = 5, 2, -7$.
The other two roots are $2$ and $-7$.

(A) Applying the column operation $C_1 \to C_1 + C_2 + C_3$,we get:
$\Delta = \begin{vmatrix} 1 + \omega^n + \omega^{2n} & \omega^n & \omega^{2n} \\ 1 + \omega^n + \omega^{2n} & 1 & \omega^n \\ 1 + \omega^n + \omega^{2n} & \omega^{2n} & 1 \end{vmatrix}$
Since $n \ne 3k$,$n$ is not a multiple of $3$. Therefore,$1 + \omega^n + \omega^{2n} = 0$.
Substituting this into the determinant:
$\Delta = \begin{vmatrix} 0 & \omega^n & \omega^{2n} \\ 0 & 1 & \omega^n \\ 0 & \omega^{2n} & 1 \end{vmatrix}$
Since all elements in the first column are $0$,the value of the determinant is $0$.

(D) Given,$\Delta = \left| \begin{array}{ccc} a & b & 0 \\ 0 & a & b \\ b & 0 & a \end{array} \right| = 0$.
Expanding the determinant along the first row,we get:
$a(a^2 - 0) - b(0 - b^2) + 0(0 - ab) = 0$
$a^3 + b^3 = 0$
$a^3 = -b^3$
Dividing both sides by $b^3$ (assuming $b \neq 0$),we get:
$\left( \frac{a}{b} \right)^3 = -1$
This implies that $\left( \frac{a}{b} \right)$ is one of the cube roots of $-1$.

(A) Let the determinant be $\Delta = \left| \begin{array}{ccc} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{array} \right|$.
Using the property $\log_a b = \frac{\ln b}{\ln a}$,we can rewrite the elements as:
$\Delta = \left| \begin{array}{ccc} 1 & \frac{\ln y}{\ln x} & \frac{\ln z}{\ln x} \\ \frac{\ln x}{\ln y} & 1 & \frac{\ln z}{\ln y} \\ \frac{\ln x}{\ln z} & \frac{\ln y}{\ln z} & 1 \end{array} \right|$.
Taking $\frac{1}{\ln x}$ common from $R_1$,$\frac{1}{\ln y}$ from $R_2$,and $\frac{1}{\ln z}$ from $R_3$ is not directly helpful,so let's multiply $R_1$ by $\ln x$,$R_2$ by $\ln y$,and $R_3$ by $\ln z$ and divide the determinant by $(\ln x \ln y \ln z)$:
$\Delta = \frac{1}{\ln x \ln y \ln z} \left| \begin{array}{ccc} \ln x & \ln y & \ln z \\ \ln x & \ln y & \ln z \\ \ln x & \ln y & \ln z \end{array} \right|$.
Since all three rows are identical,the value of the determinant is $0$.

(D) Let $A$ be the first term and $R$ be the common ratio of the $G$.$P$. Then,
$l = A R^{p-1} \Rightarrow \log l = \log A + (p-1) \log R$ ... $(i)$
$m = A R^{q-1} \Rightarrow \log m = \log A + (q-1) \log R$ ... $(ii)$
$n = A R^{r-1} \Rightarrow \log n = \log A + (r-1) \log R$ ... $(iii)$
Let the determinant be $\Delta = \left| \begin{array}{ccc} \log l & p & 1 \\ \log m & q & 1 \\ \log n & r & 1 \end{array} \right|$.
Applying the column operations $C_1 \to C_1 - (\log A) C_3$ and then $C_1 \to C_1 / (\log R)$,we observe that the first column becomes proportional to the second column shifted by a constant. More simply,applying $C_1 \to C_1 - (\log A) C_3$,we get $\log l - \log A = (p-1) \log R$,etc. Since the columns become linearly dependent (specifically,$C_1$ is a linear combination of $C_2$ and $C_3$),the value of the determinant is $0$.

(A) Let $\Delta = \left| \begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array} \right|$.
Expanding the determinant,we get:
$\Delta = a(cb - a^2) - b(b^2 - ac) + c(ba - c^2)$
$= abc - a^3 - b^3 + abc + abc - c^3$
$= 3abc - (a^3 + b^3 + c^3)$
$= -(a^3 + b^3 + c^3 - 3abc)$.
Using the algebraic identity $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$,we have:
$\Delta = -(a + b + c) \times \frac{1}{2} [(a - b)^2 + (b - c)^2 + (c - a)^2]$.
Since $a, b, c$ are positive,$(a + b + c) > 0$. Since $a, b, c$ are not all equal,at least one of $(a - b)^2, (b - c)^2, (c - a)^2$ must be positive,so the sum of squares is positive.
Therefore,$\Delta$ is negative.

(C) The given system of homogeneous linear equations is:
$x - cy - bz = 0$
$-cx + y - az = 0$
$-bx - ay + z = 0$
For this system to have a non-trivial solution (where $x, y, z$ are not all zero),the determinant of the coefficient matrix must be zero:
$\Delta = \begin{vmatrix} 1 & -c & -b \\ -c & 1 & -a \\ -b & -a & 1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1(1 - a^2) - (-c)(-c - ab) + (-b)(ac + b) = 0$
$1 - a^2 - c^2 - abc - abc - b^2 = 0$
$1 - a^2 - b^2 - c^2 - 2abc = 0$
Rearranging the terms,we get:
$a^2 + b^2 + c^2 + 2abc = 1$

(C) Since the system of equations has a non-trivial solution,the determinant of the coefficient matrix must be zero:
$\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0$
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\begin{vmatrix} a & 1 & 1 \\ 1-a & b-1 & 0 \\ 1-a & 0 & c-1 \end{vmatrix} = 0$
Expanding along the first row:
$a(b-1)(c-1) - 1(1-a)(c-1) + 1(0 - (1-a)(b-1)) = 0$
$a(b-1)(c-1) - (1-a)(c-1) - (1-a)(b-1) = 0$
Divide the entire equation by $(1-a)(1-b)(1-c)$ (noting $a, b, c \neq 1$):
$\frac{a(b-1)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(b-1)}{(1-a)(1-b)(1-c)} = 0$
$\frac{-a}{(1-a)} - \frac{1}{(1-b)} - \frac{1}{(1-c)} = 0$
$\frac{a}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0$
Since $\frac{a}{1-a} = \frac{a-1+1}{1-a} = -1 + \frac{1}{1-a}$,we substitute this into the equation:
$-1 + \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0$
$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$.

(A) Let the determinant be $\Delta$. Expanding along the first row:
$\Delta = \cos(A+B)(\cos A \cos B - \sin A \sin B) + \sin(A+B)(\sin A \cos B + \cos A \sin B) + \cos 2B(\sin A \sin A - (-\cos A \cos A)) = 0$
Using trigonometric identities:
$\Delta = \cos(A+B)\cos(A+B) + \sin(A+B)\sin(A+B) + \cos 2B(\sin^2 A + \cos^2 A) = 0$
$\Delta = \cos^2(A+B) + \sin^2(A+B) + \cos 2B(1) = 0$
$1 + \cos 2B = 0$
$\cos 2B = -1$
Since $\cos \theta = -1$ implies $\theta = (2n+1)\pi$,we have $2B = (2n+1)\pi$.
Therefore,$B = (2n+1)\frac{\pi}{2}$ for $n \in \mathbb{Z}$.

(B) Let the determinant be $D$. Expanding along the first row:
$D = \cos \theta (\cos^2 \theta + \sin^2 \theta) - \sin \theta (-\sin \theta \cos \theta + \sin \theta \cos \theta) + \cos \theta (\sin^2 \theta + \cos^2 \theta) = 0$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$D = \cos \theta (1) - \sin \theta (0) + \cos \theta (1) = 0$
$2 \cos \theta = 0$
$\cos \theta = 0$
The general solution for $\cos \theta = 0$ is $\theta = (2n + 1) \frac{\pi}{2}$,which is equivalent to $\theta = n\pi \pm \frac{\pi}{2}$.

(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: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the given coordinates $A(4, 4), B(3, -2),$ and $C(3, -16)$:
$\text{Area} = \frac{1}{2} |4(-2 - (-16)) + 3(-16 - 4) + 3(4 - (-2))|$
$\text{Area} = \frac{1}{2} |4(14) + 3(-20) + 3(6)|$
$\text{Area} = \frac{1}{2} |56 - 60 + 18|$
$\text{Area} = \frac{1}{2} |14| = 7$
Thus,the area of the triangle $ABC$ is $7$ square units.

(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the given points $(a, b + c)$,$(b, c + a)$,and $(c, a + b)$:
Area $= \frac{1}{2} |a(c + a - (a + b)) + b(a + b - (b + c)) + c(b + c - (c + a))|$
Area $= \frac{1}{2} |a(c - b) + b(a - c) + c(b - a)|$
Area $= \frac{1}{2} |ac - ab + ba - bc + cb - ca|$
Area $= \frac{1}{2} |0| = 0$.
Alternatively,using the determinant method:
Area $= \frac{1}{2} \left| \begin{matrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{matrix} \right|$
Applying $C_2 \to C_1 + C_2$:
Area $= \frac{1}{2} \left| \begin{matrix} a & a+b+c & 1 \\ b & a+b+c & 1 \\ c & a+b+c & 1 \end{matrix} \right| = \frac{a+b+c}{2} \left| \begin{matrix} a & 1 & 1 \\ b & 1 & 1 \\ c & 1 & 1 \end{matrix} \right| = 0$ (since two columns are identical).

(B) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\Delta = \frac{1}{2} \left| \begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array} \right|$.
Given that the determinants are equal,it implies that the areas of the two triangles are equal.
Equality of area does not imply that the triangles are congruent or similar.
Therefore,the correct option is $B$.

(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: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the given vertices $(1, -1)$,$(-1, 1)$,and $(-1, -1)$:
$\text{Area} = \frac{1}{2} |1(1 - (-1)) + (-1)(-1 - (-1)) + (-1)(-1 - 1)|$
$= \frac{1}{2} |1(2) + (-1)(0) + (-1)(-2)|$
$= \frac{1}{2} |2 + 0 + 2|$
$= \frac{1}{2} |4| = 2$.
Thus,the area is $2$ square units.

(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: $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the given vertices $(5, 2)$,$(2/3, 2)$,and $(-4, 3)$:
$\Delta = \frac{1}{2} |5(2 - 3) + \frac{2}{3}(3 - 2) + (-4)(2 - 2)|$
$= \frac{1}{2} |5(-1) + \frac{2}{3}(1) + (-4)(0)|$
$= \frac{1}{2} |-5 + \frac{2}{3}|$
$= \frac{1}{2} |\frac{-15 + 2}{3}|$
$= \frac{1}{2} |-\frac{13}{3}| = \frac{13}{6}$ square units.

(C) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 4$.
Substituting the given vertices $(x, 0), (1, 1), (0, 2)$:
$\frac{1}{2} |x(1 - 2) + 1(2 - 0) + 0(0 - 1)| = 4$
$\frac{1}{2} |-x + 2| = 4$
$|-x + 2| = 8$
This gives two cases:
Case $1$: $-x + 2 = 8$ $\Rightarrow -x = 6$ $\Rightarrow x = -6$
Case $2$: $-x + 2 = -8$ $\Rightarrow -x = -10$ $\Rightarrow x = 10$
Comparing with the given options,the value $x = -6$ is present.

(C) For three points to be collinear,the area of the triangle formed by them must be zero,or the determinant of the coordinates must be zero:
$\begin{vmatrix} p+1 & 1 & 1 \\ 2p+1 & 3 & 1 \\ 2p+2 & 2p & 1 \end{vmatrix} = 0$
Expanding along the first row:
$(p+1)(3 - 2p) - 1((2p+1) - (2p+2)) + 1((2p+1)(2p) - 3(2p+2)) = 0$
$(p+1)(3 - 2p) - 1(-1) + (4p^2 + 2p - 6p - 6) = 0$
$(3p - 2p^2 + 3 - 2p) + 1 + (4p^2 - 4p - 6) = 0$
$(-2p^2 + p + 3) + 1 + (4p^2 - 4p - 6) = 0$
$2p^2 - 3p - 2 = 0$
Factoring the quadratic equation:
$2p^2 - 4p + p - 2 = 0$
$2p(p - 2) + 1(p - 2) = 0$
$(2p + 1)(p - 2) = 0$
Thus,$p = 2$ or $p = -1/2$.
Given the options,the correct value is $p = 2$.

(C) For three points $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ to be collinear,the area of the triangle formed by them must be zero,or the determinant of the coordinates must be zero:
$\left| \begin{array}{ccc} 5 & 5 & 1 \\ 10 & k & 1 \\ -5 & 1 & 1 \end{array} \right| = 0$
Expanding along the first row:
$5(k - 1) - 5(10 - (-5)) + 1(10 - (-5k)) = 0$
$5k - 5 - 5(15) + 10 + 5k = 0$
$10k - 5 - 75 + 10 = 0$
$10k - 70 = 0$
$10k = 70$
$k = 7$

(A) The slope $m$ of the line passing through the origin $(0, 0)$ and the point $(x_1, y_1) = (-4, 5)$ is given by $m = \frac{y_1 - 0}{x_1 - 0} = \frac{5 - 0}{-4 - 0} = -\frac{5}{4}$.
Using the slope-intercept form $y = mx$,we get $y = -\frac{5}{4}x$.
Multiplying both sides by $4$,we get $4y = -5x$,which simplifies to $5x + 4y = 0$.

(D) Given the vector equation: $(x + 3y - 4z)i + (x - 3y + 5z)j + (3x + y)k = \lambda xi + \lambda yj + \lambda zk$.
Comparing the coefficients of $i, j,$ and $k$ on both sides, we get the following system of linear equations:
$(1 - \lambda)x + 3y - 4z = 0$ ... $(i)$
$x - (3 + \lambda)y + 5z = 0$ ... $(ii)$
$3x + y - \lambda z = 0$ ... $(iii)$
Since $(x, y, z) \ne (0, 0, 0)$, the system has a non-trivial solution, which implies the determinant of the coefficient matrix must be zero:
$\begin{vmatrix} 1 - \lambda & 3 & -4 \\ 1 & -(3 + \lambda) & 5 \\ 3 & 1 & -\lambda \end{vmatrix} = 0$.
Expanding the determinant:
$(1 - \lambda) [\lambda(3 + \lambda) - 5] - 3 [-\lambda - 15] - 4 [1 + 3(3 + \lambda)] = 0$
$(1 - \lambda) [\lambda^2 + 3\lambda - 5] + 3\lambda + 45 - 4 [1 + 9 + 3\lambda] = 0$
$(\lambda^2 + 3\lambda - 5 - \lambda^3 - 3\lambda^2 + 5\lambda) + 3\lambda + 45 - 40 - 12\lambda = 0$
$-\lambda^3 - 2\lambda^2 - \lambda = 0$
$-\lambda(\lambda^2 + 2\lambda + 1) = 0$
$-\lambda(\lambda + 1)^2 = 0$.
Wait, re-evaluating the determinant expansion:
$(1 - \lambda)(\lambda^2 + 3\lambda - 5) - 3(-\lambda - 15) - 4(1 + 9 + 3\lambda) = 0$
$(\lambda^2 + 3\lambda - 5 - \lambda^3 - 3\lambda^2 + 5\lambda) + 3\lambda + 45 - 40 - 12\lambda = 0$
$-\lambda^3 - 2\lambda^2 - \lambda = 0 \Rightarrow \lambda(\lambda^2 + 2\lambda + 1) = 0 \Rightarrow \lambda(\lambda + 1)^2 = 0$.
Checking the original equations again:
$x + 3y - 4z = \lambda x \Rightarrow (1-\lambda)x + 3y - 4z = 0$
$x - 3y + 5z = \lambda y \Rightarrow x - (3+\lambda)y + 5z = 0$
$3x + y = \lambda z \Rightarrow 3x + y - \lambda z = 0$
Determinant: $(1-\lambda)(\lambda^2+3\lambda-5) - 3(-\lambda-15) - 4(1+9+3\lambda) = 0$
$-\lambda^3 - 2\lambda^2 - \lambda + 0 = 0 \Rightarrow \lambda = 0, -1$. Thus, the values are $0, -1$.

(D) The given equations of the planes are:
$x - cy - bz = 0$
$-cx + y - az = 0$
$-bx - ay + z = 0$
For these planes to pass through a single line,the determinant of the coefficients must be zero:
$\begin{vmatrix} 1 & -c & -b \\ -c & 1 & -a \\ -b & -a & 1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1(1 - a^2) - (-c)(-c - ab) + (-b)(ac + b) = 0$
$1 - a^2 - c^2 - abc - abc - b^2 = 0$
$1 - a^2 - b^2 - c^2 - 2abc = 0$
$a^2 + b^2 + c^2 + 2abc = 1$

(D) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Alternatively,using the determinant form:
Area $= \frac{1}{2} \left| \begin{array}{ccc} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{array} \right|$
Applying the column operation $C_2 \to C_2 + C_1$:
Area $= \frac{1}{2} \left| \begin{array}{ccc} a & a+b+c & 1 \\ b & a+b+c & 1 \\ c & a+b+c & 1 \end{array} \right|$
Taking $(a+b+c)$ common from the second column:
Area $= \frac{a+b+c}{2} \left| \begin{array}{ccc} a & 1 & 1 \\ b & 1 & 1 \\ c & 1 & 1 \end{array} \right|$
Since two columns are identical,the value of the determinant is $0$.
Therefore,the area of the triangle is $0$.

(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:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the given vertices $(4, 4)$,$(3, -2)$,and $(3, -16)$:
$\text{Area} = \frac{1}{2} |4(-2 - (-16)) + 3(-16 - 4) + 3(4 - (-2))|$
$\text{Area} = \frac{1}{2} |4(14) + 3(-20) + 3(6)|$
$\text{Area} = \frac{1}{2} |56 - 60 + 18|$
$\text{Area} = \frac{1}{2} |14| = 7$
Thus,the area of the triangle is $7$ square units.

(C) For the given points to be collinear,the area of the triangle formed by them must be zero,or the determinant of the coordinates must be zero:
$\begin{vmatrix} t_1 & 2at_1 + at_1^3 & 1 \\ t_2 & 2at_2 + at_2^3 & 1 \\ t_3 & 2at_3 + at_3^3 & 1 \end{vmatrix} = 0$
Taking $a$ common from the second column:
$a \begin{vmatrix} t_1 & 2t_1 + t_1^3 & 1 \\ t_2 & 2t_2 + t_2^3 & 1 \\ t_3 & 2t_3 + t_3^3 & 1 \end{vmatrix} = 0$
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\begin{vmatrix} t_1 & 2t_1 + t_1^3 & 1 \\ t_2 - t_1 & 2(t_2 - t_1) + (t_2^3 - t_1^3) & 0 \\ t_3 - t_1 & 2(t_3 - t_1) + (t_3^3 - t_1^3) & 0 \end{vmatrix} = 0$
Expanding along the third column:
$(t_2 - t_1)(t_3 - t_1) \begin{vmatrix} 1 & 2 + (t_2^2 + t_1^2 + t_2 t_1) \\ 1 & 2 + (t_3^2 + t_1^2 + t_3 t_1) \end{vmatrix} = 0$
$(t_2 - t_1)(t_3 - t_1) [(2 + t_3^2 + t_1^2 + t_3 t_1) - (2 + t_2^2 + t_1^2 + t_2 t_1)] = 0$
$(t_2 - t_1)(t_3 - t_1) [t_3^2 - t_2^2 + t_1(t_3 - t_2)] = 0$
$(t_2 - t_1)(t_3 - t_1)(t_3 - t_2)(t_3 + t_2 + t_1) = 0$
Since $t_1, t_2, t_3$ are distinct,$(t_2 - t_1) \neq 0$,$(t_3 - t_1) \neq 0$,and $(t_3 - t_2) \neq 0$.
Therefore,$t_1 + t_2 + t_3 = 0$.

(C) Given that $x_1 = ar, x_2 = ar^2$ and $y_1 = bs, y_2 = bs^2$.
The area of the triangle $\Delta$ is given by the determinant formula:
$\Delta = \frac{1}{2} \left| \begin{matrix} a & b & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{matrix} \right| = \frac{1}{2} \left| \begin{matrix} a & b & 1 \\ ar & bs & 1 \\ ar^2 & bs^2 & 1 \end{matrix} \right|$
Taking $a$ common from the first column and $b$ common from the second column:
$\Delta = \frac{1}{2} ab \left| \begin{matrix} 1 & 1 & 1 \\ r & s & 1 \\ r^2 & s^2 & 1 \end{matrix} \right|$
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\Delta = \frac{1}{2} ab \left| \begin{matrix} 1 & 1 & 1 \\ r-1 & s-1 & 0 \\ r^2-1 & s^2-1 & 0 \end{matrix} \right|$
Expanding along the third column:
$\Delta = \frac{1}{2} ab \left| \begin{matrix} r-1 & s-1 \\ (r-1)(r+1) & (s-1)(s+1) \end{matrix} \right|$
$\Delta = \frac{1}{2} ab (r-1)(s-1) \left| \begin{matrix} 1 & 1 \\ r+1 & s+1 \end{matrix} \right|$
$\Delta = \frac{1}{2} ab (r-1)(s-1) (s+1 - r - 1) = \frac{1}{2} ab (r-1)(s-1)(s-r)$.

(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the given vertices:
$x_1 = a \cos \theta, y_1 = b \sin \theta$
$x_2 = -a \sin \theta, y_2 = b \cos \theta$
$x_3 = -a \cos \theta, y_3 = -b \sin \theta$
$\Delta = \frac{1}{2} |a \cos \theta (b \cos \theta - (-b \sin \theta)) + (-a \sin \theta) (-b \sin \theta - b \sin \theta) + (-a \cos \theta) (b \sin \theta - b \cos \theta)|$
$\Delta = \frac{1}{2} |a \cos \theta (b \cos \theta + b \sin \theta) - a \sin \theta (-2b \sin \theta) - a \cos \theta (b \sin \theta - b \cos \theta)|$
$\Delta = \frac{1}{2} |ab \cos^2 \theta + ab \sin \theta \cos \theta + 2ab \sin^2 \theta - ab \sin \theta \cos \theta + ab \cos^2 \theta|$
$\Delta = \frac{1}{2} |ab \cos^2 \theta + 2ab \sin^2 \theta + ab \cos^2 \theta|$
$\Delta = \frac{1}{2} |2ab \cos^2 \theta + 2ab \sin^2 \theta| = \frac{1}{2} |2ab(\cos^2 \theta + \sin^2 \theta)|$
Since $\cos^2 \theta + \sin^2 \theta = 1$,we get $\Delta = \frac{1}{2} |2ab| = |ab|$.

(D) Given,$D = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{array} \right|$.
Apply row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$D = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 0 & x & 0 \\ 0 & 0 & y \end{array} \right|$.
Expanding along the first column:
$D = 1 \cdot (x \cdot y - 0 \cdot 0) - 0 + 0 = xy$.
Since $D = xy$,it is clearly divisible by both $x$ and $y$.

(A) Given the matrix $A = \begin{bmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{bmatrix}$.
Since $A$ is an upper triangular matrix,its determinant $|A|$ is the product of its diagonal elements.
$|A| = 5 \times \alpha \times 5 = 25\alpha$.
We are given that $|A|^2 = 25$.
Substituting the value of $|A|$,we get $(25\alpha)^2 = 25$.
$625\alpha^2 = 25$.
$\alpha^2 = \frac{25}{625} = \frac{1}{25}$.
Taking the square root on both sides,$|\alpha| = \sqrt{\frac{1}{25}} = \frac{1}{5}$.

(D) The given system of linear equations is:
$(2-\lambda)x_1 - 2x_2 + x_3 = 0$
$2x_1 - (3+\lambda)x_2 + 2x_3 = 0$
$-x_1 + 2x_2 - \lambda x_3 = 0$
For a non-trivial solution,the determinant of the coefficient matrix must be zero:
$\begin{vmatrix} 2-\lambda & -2 & 1 \\ 2 & -(3+\lambda) & 2 \\ -1 & 2 & -\lambda \end{vmatrix} = 0$
Applying $R_1 \to R_1 + R_3$:
$\begin{vmatrix} 1-\lambda & 0 & 1-\lambda \\ 2 & -(3+\lambda) & 2 \\ -1 & 2 & -\lambda \end{vmatrix} = 0$
Taking $(1-\lambda)$ common from $R_1$:
$(1-\lambda) \begin{vmatrix} 1 & 0 & 1 \\ 2 & -(3+\lambda) & 2 \\ -1 & 2 & -\lambda \end{vmatrix} = 0$
Expanding along $R_1$:
$(1-\lambda) [1(\lambda(3+\lambda) - 4) + 1(4 - (3+\lambda))] = 0$
$(1-\lambda) [3\lambda + \lambda^2 - 4 + 4 - 3 - \lambda] = 0$
$(1-\lambda) [\lambda^2 + 2\lambda - 3] = 0$
$(1-\lambda)(\lambda+3)(\lambda-1) = 0$
$-(1-\lambda)^2(\lambda+3) = 0$
Thus,$\lambda = 1$ and $\lambda = -3$. The set of values is $\{1, -3\}$,which contains two elements.

(B) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$\Delta = 0$.
The coefficient matrix is:
$\Delta = \begin{vmatrix} 1 & \lambda & -1 \\ \lambda & -1 & -1 \\ 1 & 1 & -\lambda \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1((-1)(-\lambda) - (-1)(1)) - \lambda((\lambda)(-\lambda) - (-1)(1)) - 1((\lambda)(1) - (-1)(1)) = 0$
$1(\lambda + 1) - \lambda(-\lambda^2 + 1) - 1(\lambda + 1) = 0$
$(\lambda + 1) - \lambda(1 - \lambda^2) - (\lambda + 1) = 0$
$-\lambda(1 - \lambda)(1 + \lambda) = 0$
$\lambda(1 - \lambda)(1 + \lambda) = 0$
Solving for $\lambda$,we get $\lambda = 0, 1, -1$.
Thus,there are exactly three values of $\lambda$ for which the system has a non-trivial solution.

(D) Let the third vertex be $(p, q)$. Since it lies on the line $y = x + 3$,we have $q = p + 3$ $(i)$.
Given the area of the triangle is $5$,the area formula is:
$\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 5$
$\frac{1}{2} |2(-2 - q) + 3(q - 1) + p(1 - (-2))| = 5$
$|2(-2 - q) + 3(q - 1) + 3p| = 10$
$|-4 - 2q + 3q - 3 + 3p| = 10$
$|q + 3p - 7| = 10$
Substituting $q = p + 3$ into the equation:
$|(p + 3) + 3p - 7| = 10$
$|4p - 4| = 10$
Case $1$: $4p - 4 = 10$ $\Rightarrow 4p = 14$ $\Rightarrow p = \frac{7}{2}$. Then $q = \frac{7}{2} + 3 = \frac{13}{2}$.
Case $2$: $4p - 4 = -10$ $\Rightarrow 4p = -6$ $\Rightarrow p = -\frac{3}{2}$. Then $q = -\frac{3}{2} + 3 = \frac{3}{2}$.
The possible vertices are $\left( \frac{7}{2}, \frac{13}{2} \right)$ or $\left( -\frac{3}{2}, \frac{3}{2} \right)$.
Comparing with the given options,the correct option is $D$.

(C) Given vertices are $(a, b)$,$(x_1, y_1) = (ar, bs)$,and $(x_2, y_2) = (ar^2, bs^2)$.
The area of the triangle is given by the determinant formula:
$\Delta = \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$
Substituting the values:
$\Delta = \frac{1}{2} |a(bs - bs^2) + ar(bs^2 - b) + ar^2(b - bs)|$
Factoring out $ab$:
$\Delta = \frac{1}{2} ab |(s - s^2) + r(s^2 - 1) + r^2(1 - s)|$
$= \frac{1}{2} ab |s(1 - s) + r(s - 1)(s + 1) - r^2(s - 1)|$
$= \frac{1}{2} ab |(s - 1) [-s + r(s + 1) - r^2]|$
$= \frac{1}{2} ab |(s - 1) [-s + rs + r - r^2]|$
$= \frac{1}{2} ab |(s - 1) [r(s - r) - (s - r)]|$
$= \frac{1}{2} ab |(s - 1)(s - r)(r - 1)|$
$= \frac{1}{2} ab (r - 1)(s - 1)(s - r)$.

(D) Given $A = \begin{vmatrix} \sin(\theta + \alpha) & \cos(\theta + \alpha) & 1 \\ \sin(\theta + \beta) & \cos(\theta + \beta) & 1 \\ \sin(\theta + \gamma) & \cos(\theta + \gamma) & 1 \end{vmatrix}$.
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$A = \begin{vmatrix} \sin(\theta + \alpha) & \cos(\theta + \alpha) & 1 \\ \sin(\theta + \beta) - \sin(\theta + \alpha) & \cos(\theta + \beta) - \cos(\theta + \alpha) & 0 \\ \sin(\theta + \gamma) - \sin(\theta + \alpha) & \cos(\theta + \gamma) - \cos(\theta + \alpha) & 0 \end{vmatrix}$.
Expanding along the third column:
$A = 1 \cdot [(\sin(\theta + \beta) - \sin(\theta + \alpha))(\cos(\theta + \gamma) - \cos(\theta + \alpha)) - (\cos(\theta + \beta) - \cos(\theta + \alpha))(\sin(\theta + \gamma) - \sin(\theta + \alpha))]$.
Using the identity $\sin(x)\cos(y) - \cos(x)\sin(y) = \sin(x-y)$:
$A = \sin((\theta + \beta) - (\theta + \gamma)) + \sin((\theta + \alpha) - (\theta + \beta)) + \sin((\theta + \gamma) - (\theta + \alpha))$.
$A = \sin(\beta - \gamma) + \sin(\alpha - \beta) + \sin(\gamma - \alpha)$.
Since the result contains only constants $\alpha, \beta, \gamma$,$A$ is independent of $\theta$.

(A) Given determinant is $f(x) = \left| \begin{array}{ccc} 1 & x & x + 1 \\ 2x & x(x - 1) & (x + 1)x \\ 3x(x - 1) & x(x - 1)(x - 2) & (x + 1)x(x - 1) \end{array} \right|$.
Taking $x$ common from $C_2$ and $(x+1)$ common from $C_3$ is not directly possible,but we can factor out $x$ from the second column and $x(x+1)$ from the third row.
Alternatively,observe the columns. Notice that $C_3 = C_2 + C_1$ is not immediately obvious,but let's simplify by taking common factors:
$f(x) = x(x-1) \cdot x \cdot \left| \begin{array}{ccc} 1 & 1 & x+1 \\ 2x & x-1 & x(x+1) \\ 3x(x-1) & (x-2) & (x+1)x \end{array} \right|$.
Actually,a simpler way is to notice that for any $x$,the rows are linearly dependent. Let $R_3 = R_3 - (x-2)R_2$.
Performing row operations or expanding the determinant shows that the expression simplifies to $0$ for all $x$.
Specifically,$f(x) = 0$ for all $x \in \mathbb{R}$.
Therefore,$f(100) = 0$.

(C) Let $\Delta = \left| \begin{array}{ccc} 10! & 11! & 12! \\ 11! & 12! & 13! \\ 12! & 13! & 14! \end{array} \right|$.
Taking $10!$ common from $C_1$,$11!$ from $C_2$,and $12!$ from $C_3$:
$\Delta = 10!\,11!\,12! \left| \begin{array}{ccc} 1 & 11 & 11 \times 12 \\ 1 & 12 & 12 \times 13 \\ 1 & 13 & 13 \times 14 \end{array} \right|$.
Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\Delta = 10!\,11!\,12! \left| \begin{array}{ccc} 1 & 11 & 132 \\ 0 & 1 & 156-132 \\ 0 & 2 & 182-132 \end{array} \right| = 10!\,11!\,12! \left| \begin{array}{ccc} 1 & 11 & 132 \\ 0 & 1 & 24 \\ 0 & 2 & 50 \end{array} \right|$.
Expanding along $C_1$:
$\Delta = 10!\,11!\,12! \times (1 \times (50 - 48)) = 10!\,11!\,12! \times 2 = 2(10!\,11!\,12!)$.

(A) The given determinant is $\Delta = \left| \begin{array}{ccc} \cos(A-P) & \cos(A-Q) & \cos(A-R) \\ \cos(B-P) & \cos(B-Q) & \cos(B-R) \\ \cos(C-P) & \cos(C-Q) & \cos(C-R) \end{array} \right|$.
Using the formula $\cos(x-y) = \cos x \cos y + \sin x \sin y$,we can write each element as a sum of two terms.
This allows us to express the determinant as the product of two matrices:
$\Delta = \left| \begin{array}{ccc} \cos A & \sin A & 0 \\ \cos B & \sin B & 0 \\ \cos C & \sin C & 0 \end{array} \right| \times \left| \begin{array}{ccc} \cos P & \cos Q & \cos R \\ \sin P & \sin Q & \sin R \\ 0 & 0 & 0 \end{array} \right|$.
Since the third column of the first matrix is all zeros and the third row of the second matrix is all zeros,the product of these two matrices is zero.
Alternatively,the determinant can be split into $8$ determinants,each of which has at least two identical columns (or rows),making the value of each determinant $0$.
Thus,the value of the determinant is $0$.

(C) Let the given determinant be $\Delta$. Adding $R_2$ and $R_3$ to $R_1$,we get:
$\Delta = \left| {\begin{array}{*{20}{c}}{\sin x + 2\cos x}&{\sin x + 2\cos x}&{\sin x + 2\cos x}\\{\cos x}&{\sin x}&{\cos x}\\{\cos x}&{\cos x}&{\sin x}\end{array}} \right| = 0$
Taking $(\sin x + 2\cos x)$ common from $R_1$:
$\Delta = (\sin x + 2\cos x) \left| {\begin{array}{*{20}{c}}1&1&1\\{\cos x}&{\sin x}&{\cos x}\\{\cos x}&{\cos x}&{\sin x}\end{array}} \right| = 0$
Applying $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:
$\Delta = (\sin x + 2\cos x) \left| {\begin{array}{*{20}{c}}1&0&0\\{\cos x}&{\sin x - \cos x}&0\\{\cos x}&0&{\sin x - \cos x}\end{array}} \right| = 0$
$\Delta = (\sin x + 2\cos x)(\sin x - \cos x)^2 = 0$
This implies $\sin x + 2\cos x = 0$ or $(\sin x - \cos x)^2 = 0$.
Case $1$: $\tan x = -2$. In the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$,$\tan x$ ranges from $-1$ to $1$. Thus,$\tan x = -2$ has no solution.
Case $2$: $\sin x = \cos x \implies \tan x = 1$. In the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$,$\tan x = 1$ at $x = \frac{\pi}{4}$.
Thus,there is only $1$ distinct real root.

(D) Given the determinant $\Delta = \begin{vmatrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{vmatrix}$.
Expanding along the first row:
$\Delta = C(C^2 - 1) - 1(C - 6) + 0(1 - 6C)$
$\Delta = C^3 - C - C + 6$
$\Delta = C^3 - 2C + 6$
Substitute $C = 2 \cos \theta$ into the expression:
$\Delta = (2 \cos \theta)^3 - 2(2 \cos \theta) + 6$
$\Delta = 8 \cos^3 \theta - 4 \cos \theta + 6$
Since $4 \cos^3 \theta - 3 \cos \theta = \cos 3\theta$,we have $8 \cos^3 \theta = 2 \cos 3\theta + 6 \cos \theta$.
Thus,$\Delta = 2 \cos 3\theta + 6 \cos \theta - 4 \cos \theta + 6 = 2 \cos 3\theta + 2 \cos \theta + 6$.
None of the given options match this result.

(D) For the matrix to be singular,its determinant must be zero: $\begin{vmatrix} 3 & -1+x & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2 \end{vmatrix} = 0$.
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\begin{vmatrix} 3 & -1+x & 2 \\ 0 & -x & x \\ x & -x & 0 \end{vmatrix} = 0$.
Applying column operation $C_1 \to C_1 + C_2 + C_3$:
$\begin{vmatrix} x+4 & -1+x & 2 \\ 0 & -x & x \\ 0 & -x & 0 \end{vmatrix} = 0$.
Expanding along the first column:
$(x+4) \cdot [(-x)(0) - (x)(-x)] = 0$
$(x+4)(x^2) = 0$.
This gives $x = -4$ or $x = 0$.
We are given the interval $[-4, -1]$.
Since $-4 \in [-4, -1]$ and $0 \notin [-4, -1]$,there is only $1$ value of $x$ that satisfies the condition.

(C) For the system of linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero.
The coefficient matrix is:
$\Delta = \begin{vmatrix} a^3 & (a+1)^3 & (a+2)^3 \\ a & a+1 & a+2 \\ 1 & 1 & 1 \end{vmatrix} = 0$
Applying column operations $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_2$:
$\Delta = \begin{vmatrix} a^3 & (a+1)^3 - a^3 & (a+2)^3 - (a+1)^3 \\ a & 1 & 1 \\ 1 & 0 & 0 \end{vmatrix} = 0$
Expanding along the third row $(R_3)$:
$1 \cdot \begin{vmatrix} (a+1)^3 - a^3 & (a+2)^3 - (a+1)^3 \\ 1 & 1 \end{vmatrix} = 0$
$((a+1)^3 - a^3) - ((a+2)^3 - (a+1)^3) = 0$
$(3a^2 + 3a + 1) - (3(a+1)^2 + 3(a+1) + 1) = 0$
$(3a^2 + 3a + 1) - (3a^2 + 6a + 3 + 3a + 3 + 1) = 0$
$(3a^2 + 3a + 1) - (3a^2 + 9a + 7) = 0$
$-6a - 6 = 0$
$-6a = 6$
$a = -1$.

(C) Given the identity $px^4 + qx^3 + rx^2 + sx + t = \left| \begin{array}{ccc} x^2 + 3x & x - 1 & x + 3 \\ x + 1 & 2 - x & x - 3 \\ x - 3 & x + 4 & 3x \end{array} \right|$.
To find the value of $t$,we set $x = 0$ in the identity.
Substituting $x = 0$ on both sides:
$p(0)^4 + q(0)^3 + r(0)^2 + s(0) + t = \left| \begin{array}{ccc} 0^2 + 3(0) & 0 - 1 & 0 + 3 \\ 0 + 1 & 2 - 0 & 0 - 3 \\ 0 - 3 & 0 + 4 & 3(0) \end{array} \right|$
$t = \left| \begin{array}{ccc} 0 & -1 & 3 \\ 1 & 2 & -3 \\ -3 & 4 & 0 \end{array} \right|$
Now,expand the determinant along the first row:
$t = 0(2 \times 0 - (-3) \times 4) - (-1)(1 \times 0 - (-3) \times (-3)) + 3(1 \times 4 - 2 \times (-3))$
$t = 0 - (-1)(0 - 9) + 3(4 + 6)$
$t = 0 - (9) + 3(10)$
$t = -9 + 30$
$t = 21$
Thus,the value of $t$ is $21$.

(A) Given the determinant $\Delta = \left| \begin{array}{ccc} a+1 & a+2 & a+p \\ a+2 & a+3 & a+q \\ a+3 & a+4 & a+r \end{array} \right| = 0$.
Apply row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_2$:
$\Delta = \left| \begin{array}{ccc} a+1 & a+2 & a+p \\ 1 & 1 & q-p \\ 1 & 1 & r-q \end{array} \right| = 0$.
Since two rows ($R_2$ and $R_3$) have identical elements in the first two columns,we can perform $R_3 \rightarrow R_3 - R_2$:
$\Delta = \left| \begin{array}{ccc} a+1 & a+2 & a+p \\ 1 & 1 & q-p \\ 0 & 0 & (r-q) - (q-p) \end{array} \right| = 0$.
Expanding along the third row:
$0 - 0 + (r - q - q + p) \times ((a+1)(1) - (a+2)(1)) = 0$.
$(r + p - 2q) \times (-1) = 0$.
This implies $r + p - 2q = 0$,or $p + r = 2q$.
This is the condition for $p, q, r$ to be in $AP$.

(C) Given $a^2 + b^2 + c^2 = -2$.
Apply the column operation $C_1 \rightarrow C_1 + C_2 + C_3$:
$f(x) = \left| \begin{array}{ccc} 1 + x(a^2 + b^2 + c^2 + 2) & (1 + b^2)x & (1 + c^2)x \\ 1 + x(a^2 + b^2 + c^2 + 2) & 1 + b^2x & (1 + c^2)x \\ 1 + x(a^2 + b^2 + c^2 + 2) & (1 + b^2)x & 1 + c^2x \end{array} \right|$
Since $a^2 + b^2 + c^2 = -2$,the term $a^2 + b^2 + c^2 + 2 = 0$.
Thus,$f(x) = \left| \begin{array}{ccc} 1 & (1 + b^2)x & (1 + c^2)x \\ 1 & 1 + b^2x & (1 + c^2)x \\ 1 & (1 + b^2)x & 1 + c^2x \end{array} \right|$
Apply row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$f(x) = \left| \begin{array}{ccc} 1 & (1 + b^2)x & (1 + c^2)x \\ 0 & 1 - x & 0 \\ 0 & 0 & 1 - x \end{array} \right|$
Expanding along the first column:
$f(x) = 1 \cdot (1 - x)(1 - x) = (1 - x)^2 = 1 - 2x + x^2$.
Since the highest power of $x$ is $2$,$f(x)$ is a polynomial of degree $2$.

(D) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.
The determinant is given by:
$\Delta = \begin{vmatrix} \sin \theta & -\cos \theta & \lambda + 1 \\ \cos \theta & \sin \theta & -\lambda \\ \lambda & \lambda + 1 & \cos \theta \end{vmatrix} = 0$
Expanding the determinant along the first row:
$\Delta = \sin \theta (\sin \theta \cos \theta - (-\lambda)(\lambda + 1)) - (-\cos \theta)(\cos^2 \theta - (-\lambda)(\lambda)) + (\lambda + 1)(\cos \theta (\lambda + 1) - \lambda \sin \theta) = 0$
Simplifying the expression:
$\Delta = \sin \theta (\sin \theta \cos \theta + \lambda^2 + \lambda) + \cos \theta (\cos^2 \theta + \lambda^2) + (\lambda + 1)(\lambda \cos \theta + \cos \theta - \lambda \sin \theta) = 0$
After algebraic simplification,the equation reduces to:
$2 \cos \theta (\lambda^2 + \lambda + 1) = 0$
Since $\lambda^2 + \lambda + 1 = (\lambda + \frac{1}{2})^2 + \frac{3}{4} > 0$ for all $\lambda \in \mathbb{R}$,we must have:
$\cos \theta = 0$
Therefore,$\theta = (2n + 1)\frac{\pi}{2}$ for $n \in \mathbb{I}$.