$\displaystyle \begin{bmatrix} 3x+7 & 5

\ y+1 & 2-3x \end{bmatrix}=\begin{bmatrix} 0 & y-2 \ 8 & 4

\end{bmatrix}$
Equating the corresponding elements, we get,
$\displaystyle 3x+7=0\Rightarrow x=-\frac { 7 }{ 3 } $
$\displaystyle 5=y-2\Rightarrow y=7$
$\displaystyle y+1=8\Rightarrow y=7$
$\displaystyle 2-3x=4\Rightarrow x=-\frac { 2 }{ 3 } $
We find that on comparing the corresponding elements of the two matrices, we get two different values of $x$, which is not possible.
Hence, it is not possible to find the values of $x$ and $y$ for which the given matrices are equal.

$kx-ay=66-----(1)$

Homogeneous steamy has non- trivial solution it's mean determinant is zero

For infinitely many solutions,
$\displaystyle \frac { k+1 }{ k } =\frac { 8 }{ k+3 } =\frac { 4 }{ 3k-1 } $

$\Rightarrow k^{2}-4k+3=0 $ and $24k-8=4k+12$
$\Rightarrow k=3,k=1$ and $k=1$

Hence, $k=1$

$\left|\begin{matrix} a^2 & a & 1 \ \sin(n+1)x & \sin{nx} & \sin(n-1)x \ \cos(n+1)x & \cos{nx} & \cos(n-1)x\end{matrix}\right| = 0$
 
$ \Rightarrow \displaystyle { a }^{ 2 }\left[ \sin { nx } .\cos { \left( n-1 \right) x } -\cos { nx } .\sin { \left( n-1 \right) x }  \right] +a\left[ \sin  (n-1)x.\cos  (n+1)x-\cos  (n-1)x.\sin  (n+1)x \right] \ +1\left[ \sin  (n+1)x.\cos { nx } -\cos  (n+1)x.\sin { nx }  \right] =0$

$ \displaystyle \Rightarrow a^{ 2 }\sin  x-a\sin  2x+\sin  x=0\ \displaystyle \Rightarrow \sin  x\left( { a }^{ 2 }+1-2a\cos { x }  \right) =0\ \displaystyle \Rightarrow \sin  x=0\; { or }\; \cos { x } =\frac { { a }^{ 2 }+1 }{ 2a } \ \displaystyle \Rightarrow \sin  x=0\; { or }\; \cos { x } =1\quad \left( \because { a }^{ 2 }+1\ge 2a,\; a>0\Rightarrow \frac { { a }^{ 2 }+1 }{ 2a } \ge 1 \right) \ \displaystyle \Rightarrow x=n\pi \; or\; x=2n\pi, \; n \in I \ \displaystyle\Rightarrow x=n\pi $

Given equation of lines passes through same point i.e. lines are concurrent
$kx-2y+3=0\ 
x+ky+2=0\ 
2x+k=0$
$\Rightarrow \left| \begin{matrix} 2k & -2 & 3 \ 1 & k & 2 \ 2 & 0 & k \end{matrix} \right| =0$
$\Rightarrow k^{3}-2k-4=0$
$\Rightarrow (k-2)(k^2+2k+2)=0$
The discriminant of the quadratic expression is negative.
Hence there is only one real value of $k$

$\Delta =\begin{vmatrix} k & 1 & 1 \ 1 & k & 1 \ 1 & 1 & k \end{vmatrix}$

$\Rightarrow \Delta =(k-1)^{2}(k+2)$

Taking $\Delta=0$
$\Rightarrow (k-1)^{2}(k+2)=0$
$\Rightarrow k=1, k=-2$
At these values of k, system can have either no solution or infinitely many solution.

For k=1, equations takes the form $x+y+z=1$
Hence, infinitely many solution for k=1.

For k=-2,
$D _{1}=\begin{vmatrix} 1 & 1 & 1 \ -2 & -2 & 1 \ 4 & 1 & -2 \end{vmatrix}$
$D _{1}=9 \ne 0$

So, $D=0$, at least one $D _{1}\ne 0$
Hence, system has no solution for k=-2

Given system of equations can be written as
$AX=B$

$\begin{bmatrix} 1 & 1 & 1 \ 2 & -1 & 3 \ 1 & -2 & -1 \end{bmatrix}: \begin{bmatrix} x \ y \ z \end{bmatrix}=\begin{bmatrix} 2 \ 5 \ -1 \end{bmatrix}$

For non trivial solution

$ax+by+c=0,bx+cy+a=0,cx+ay+b=0$
Gives
$\begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix}=0$
Now $(b+c)x+(c+a)y+(a+b)z=0,(c+a)x+(a+b)y+(b+c)z=0,(a+b)x+(b+c)y+(c+a)z=0$
gives
$\begin{vmatrix} b+c & c+a & a+b \ c+a & a+b & b+c \ a+b & b+c & c+a \end{vmatrix}=\begin{vmatrix} b & c & a \ c & a & b \ a & b & c \end{vmatrix}+\begin{vmatrix} c & a & b \ a & b & c \ b & c & a \end{vmatrix}=\begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix}+\begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix}=2\begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix}=0$

To   find  all  values   of    $\alpha$   which   are  possible
For  $\lambda = 1 $
$ \begin{bmatrix}
1  & sin\alpha  & cos\alpha \
 1 & cos\alpha  & sin\alpha  \
 -1& sin\alpha   &  cos\alpha 
\end{bmatrix} = \begin{bmatrix}
0\
0\
0\end{bmatrix}$

On   Solving   this   equation, we   get
$ 2cos2\alpha = 0$
$ cos2\alpha  = 0 $
$ 2\alpha  = \dfrac{\pi }{2} $
 Finally  , we  get

$ \alpha  = \dfrac{\pi }{8}   $ 
 
$ \alpha  = \dfrac{7\pi }{8}   $     and

$ \alpha  = \dfrac{ 9\pi }{8}   $  

Here, $D =\begin{vmatrix} \alpha & 1 & 1 \ 1 & \alpha & 1 \ 1 & 1 & \alpha \end{vmatrix}$

$\Rightarrow D =(\alpha-1)^{2}(\alpha+2)$

Taking $D=0$
$\Rightarrow (\alpha-1)^{2}(\alpha+2)=0$
$\Rightarrow \alpha=1, \alpha=-2$
At these values of $\alpha$, system can have either no solution or infinitely many solution.

For $\alpha=1$, equations takes the form $x+y+z=0$
Hence, infinitely many solution for $\alpha=1$.

For $\alpha=-2$,
$D _{1}=\begin{vmatrix} -3 & 1 & 1 \ -3 & -2 & 1 \ -3 & 1 & -2 \end{vmatrix}$
$D _{1}=27 \ne 0$

So, $D=0$, at least one $D _{1}\ne 0$
Hence, system has no solution for $\alpha=-2$

If the given matrices are equal then we have the following equations,

Applying $\displaystyle R _{1}\rightarrow R _{1}+R _{2}+R _{3}$ we get 

$(b+c)(y+z)-ax=b-c\$       ..............(1)

Given system of equations can be written as
$AX=B$
where $A=\begin{bmatrix} { 1 } & { 1 } & { 1 } \ { 1 } & { 2 } & 3\ { 2 } & 3 & 4 \end{bmatrix}$ 
$X=\begin{bmatrix} x \ y \ z \end{bmatrix}$ ;$B=\begin{bmatrix} 3 \ 4 \ 7 \end{bmatrix}$

Here, $|A|=0$
Now, we will find $(adj A)B$

$adj A=C^{T}={\begin{bmatrix} { -1 } & { 2 } & { -1 } \ { -1 } & { 2 } & -1 \ { 1 } & -2 & 1 \end{bmatrix}}^T$

$\Rightarrow adj A=\begin{bmatrix} { -1 } & { -1 } & { 1 } \ { 2 } & { 2 } & -2 \ { -1 } & -1 & 1 \end{bmatrix}$

Now, $(adj A)B=\begin{bmatrix} { -1 } & { -1 } & { 1 } \ { 2 } & { 2 } & -2 \ { -1 } & -1 & 1 \end{bmatrix}\begin{bmatrix} 3 \ 4 \ 7 \end{bmatrix}$

$\Rightarrow (adj A)B=\begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}$
$\Rightarrow (adj A)B=O$

Hence,the system of equations has infinitely many solutions.
Let $z=k$ where $k\in R$
Then 
$x+y=3-k$
$x+2y=4-3k$
Solving these eqns, we get 
$y=1-2k ; x=2+k$

$\Delta =\begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 2 & \lambda  \end{vmatrix}=1\left( 2\lambda -6 \right) -1\left( \lambda -3 \right) +1\left( 2-2 \right) =2\lambda -6-\lambda +3+0=\lambda -3\ \Delta _{ 1 }=\begin{vmatrix} 6 & 1 & 1 \ 10 & 2 & 3 \ \mu  & 2 & \lambda  \end{vmatrix}=6\left( 2\lambda -6 \right) -1\left( 10\lambda -3\mu  \right) +1\left( 20-2\mu  \right) \ =12\lambda -36-10\lambda +3\mu +20-2\mu =2\lambda +\mu -16$

$\Delta=\begin{vmatrix} 1 &4  &-2  \3  &1  &5  \  2&3  &1  \end{vmatrix}=1(1-15)-4(3-10)-2(9-2)=0$
and $\Delta _1$ is not zero
Therefore, it has no solution

For $x,y,z$ not to be simultaneously zero

$2x+3y+5z=9$,

${2}x+{3}y+z={5}$

The lines are concurrent,
$=>\begin{bmatrix}a&b& c\ b &c &a\ c & a& b\end{bmatrix}=0$
Applying,$ C _{1}=C _{1}+C _{2}+C _{3}$,
$=\begin{bmatrix}a+b+c&b& c\ a+b+c &c &a\ a+b+c & a& b\end{bmatrix}$
Applying, $R _{1}=R _{1}-R _{2}$ and $R _{2}=R _{2}-R _{3}$,
$=(a+b+c)\begin{bmatrix}0&b-c& c-a\ 0 &c-a &a-b\ 1 & a& b\end{bmatrix}$
Expanding by $C _{1}$,
$=(a+b+c)(-a^{2}-b^{2}-c^{2}+ab+bc+ac)=0$
$=> (a+b+c)=0$
or,
$(a+b+c)(-a^{2}-b^{2}-c^{2}+ab+bc+ac)=0$
$abc-a^{3}-b^{3}+abc+abc-c^{3}=0$
$a^{3}+b^{3}+c^{3}=3abc$
So options are A and B.

Given $\omega$ is a cube root of unity and $x+ y + z = a, x + \omega y + \omega^2 z = b, x + \omega^2 y + \omega z = c$.

$f(x) = ax^2 + bx + c; a, b, c \in  R$ and equation $f(x)- x = 0$ has imaginary roots $\alpha, \beta$ and $\gamma, \delta$ be the roots of $f(f(x)) - x = 0$
since $\alpha,\beta$ are roots of $f(x)- x = 0$
$f(\alpha)-\alpha =0$ $\Rightarrow f(\alpha)=\alpha$
$f(\beta)-\beta=0$ $\Rightarrow f(\beta)=\beta$
$f(f(\alpha))-\alpha = f(\alpha)-\alpha =0$
$f(f(\beta))-\beta = f(\beta)-\beta =0$
$\therefore  \alpha,\beta$ are also roots of $f(f(x)) - x = 0$ ------(*)
$f(x)- x= ax^2+(b-1)x+c=0$
roots are imaginary.
i.e $\alpha,\beta$ are conjugate to each other and $D<0$
$\Rightarrow (b-1)^2-4ac<0$ -------(1)
$f(f(x)) - x = a(ax^2+bx+c)^2+b(ax^2+bx+c)+c-x=0$
$\Rightarrow \left(ax^2+(b-1)x+c\right)\left(a^2x^2+(ab+a)x+ac+b+1\right)=0$
$D=(ab+a)^2-4a^2(ac+b+1) = a^2\left((b-1)^2-4ac\right)-4a^2<0$    ($\because$ from (1))
$\therefore \gamma,\delta$ are also imaginary roots and conjugate to each other.
$\begin{vmatrix} 2 & \alpha & \delta\ \beta & 0 & \alpha\ \gamma & \beta & 1\end{vmatrix} $ $= -3\alpha\beta+\alpha^2\gamma+\beta^2\delta$ ------ (2)
$\alpha\beta$ is real
$\alpha^2\gamma$ is conjugate to $\beta^2\delta$
$\Rightarrow \alpha^2\gamma+\beta^2\delta$ is real
from (2)
$\therefore \begin{vmatrix} 2 & \alpha & \delta\ \beta & 0 & \alpha\ \gamma & \beta & 1\end{vmatrix} $ is purely real.
Hence, option B.

Given $x+y+z=a $...(i)

For the solution of the system of equations
$x - 2y + 3z = -1$
$-x + y - 2z = 2$
$kx - 3y + 4z = 1$, 
$\begin{bmatrix}
1 &-2  &3 \ 
-1 &1  &-2 \ 
k &-3  &4 
\end{bmatrix} 
$
If determinant of the matrix is zero than the given lines are coplanar and if the determinant is non zero the they are non co-planar.
The solution for the system of equations exist when the lines are coplanar.
Hence, $1(-2 - 4k) - 3(-1 - k) -1(-4 + 2) \neq 0$
$\Rightarrow -2 - 4k + 3 + 3k + 2 \neq 0$
i.e. $k \neq 3$

$\begin{vmatrix} 1+\sin ^{ 2 } \theta  & \cos ^{ 2 } \theta  & 4\sin  6\theta  \ \sin ^{ 2 } \theta  & 1+\cos ^{ 2 } \theta  & 4\sin  6\theta  \ \sin ^{ 2 } \theta  & \cos ^{ 2 } \theta  & 1+4\sin  6\theta  \end{vmatrix}=0$

Applying $\displaystyle R _{1}\rightarrow R _{1}+R _{2}+R _{3}$ and taking $\displaystyle a+b+c-x $ common from the first row, we obtain
$\displaystyle \Delta =\left ( a+b+c-x \right )\begin{vmatrix}
1 &1  &1 \ 
c &b-x  &a \ 
b &a  &c-x 
\end{vmatrix}$
Applying  $\displaystyle C _{2}\rightarrow C _{2}-C _{1}$ and $\displaystyle C _{3}\rightarrow C _{3}-C _{1}$ we obtain
$\displaystyle \Delta =\begin{vmatrix}
1 &0  &0 \ 
c &b-c-x  &a-c \ 
b &a-b  &c-b-x 
\end{vmatrix}\left [ \because a+b+c=0 \right ]$
Expanding along $\displaystyle R _{1}$ we get
$\displaystyle \Delta =x\left [ \left ( b-c-x \right )\left (c-b-x  \right )-\left ( a-b \right )\left ( a-c \right ) \right ] $
$\displaystyle \Delta =x\left [ \left ( a-b \right )\left (a-c  \right )-\left ( x+b-c\right )\left ( x-b+c \right ) \right ] $
$\displaystyle  =x\left [ a^{2}-ab-ac+bc-x^{2}+b^{2}+c^{2}-2bc \right ]$
$\displaystyle \Delta =x\left [ a^{2}+b^{2}+c^{2}-bc-ab-ac-x^{2} \right ]$
$\displaystyle \Delta =0$ implies $\displaystyle x =0$ or $\displaystyle x^{2}=a^{2}+b^{2}+c^{2}-bc-ab-ac$
Now $\displaystyle x^{2}=a^{2}+b^{2}+c^{2}-bc-ab-ac$
$\displaystyle =a^{2}+b^{2}+c^{2}-\frac{1}{2}\left [ \left ( a+b+c \right )^{2}- a^{2}-b^{2}-c^{2} \right ]$
$\displaystyle =\frac{3}{2}\left ( a^{2}+b^{2}+c^{2} \right )\left [ \because a+b+c=0 \right ]$
$\displaystyle \Rightarrow x= \pm \sqrt{\frac{3}{2}\left ( a^{2}+b^{2}+c^{2} \right )}$

$D=\begin{vmatrix}1 &1  &1 \ 2 & 2 & 2\ 3 & 3 & 3\end{vmatrix}=0$
$D _1=0,$     $D _2=0$,      $D _3=0$
Let $z=t$
$x+y=1-t$
$2x+2y=3-2t$
Since both the lines are parallel hence no value of x and y
Hence there is no solution of the given equations.

We find $D = 0$, where $D$ is the determinant formed by the coefficients of $x,\ y,\ z$ in the three equations and since no pair of planes are parallel, so there is an infinite number of solutions.
Let $\alpha P _{1} + \lambda P _{2} = P _{3}$
$\Rightarrow P _{1} + 7P _{2} = 13P _{3}$
$\Rightarrow b _{1} + 7b _{2} = 13b _{3}$
(A) $D\neq 0\Rightarrow$ unique solution for any $b _{1}, b _{2}, b _{3}$
(B) $D = 0$ but $P _{1} + 7P _{2} \neq 13P _{3}$
(C) $D = 0$ Also $b _{2} = -2b _{1}, b _{3} = -b _{1}$
Satisfied $b _{1} + 7b _{2} = 13b _{3}$ (Actually all three planes are co-incident)
(D) $D\neq 0$.

For non-trivial solutions
$\begin{vmatrix}
1 &2  &1 \ 
2 &3  &-1 \ 
tan\theta &1  &-3 
\end{vmatrix}=0$
$\Rightarrow 6-5\tan\theta =0$
$\Rightarrow\displaystyle \tan \theta=\frac{6}{5}$
Hence, the number of solutions in $(-\pi,\pi)$ is 2
Hence, option 'C' is correct.