(D)

Given, $x+\left(\sqrt{2}\sin \alpha \right)y+\left(\sqrt{2}\cos \alpha \right)z=0$

$x+\left(\cos \alpha \right)y+\left(\sin \alpha \right)z=0$

$x+\left(\sin \alpha \right)y-\left(\cos \alpha \right)z=0$

For non-trivial solution,

$\left|\begin{array}{ccc}1& \sqrt{2}\sin \alpha & \sqrt{2}\cos \alpha \\ 1& \cos \alpha & \sin \alpha \\ 1& \sin \alpha & -\cos \alpha \end{array}\right|$$=0$

$\Rightarrow 1[-{\cos }^2\alpha -{\sin }^2\alpha ]-1[-\sqrt{2}\sin \alpha \cos \alpha -\sqrt{2}\sin \alpha \cos \alpha ]+1[\sqrt{2}{\sin }^2\alpha -\sqrt{2}{\cos }^2\alpha ]=0$

$\Rightarrow -1+2\sqrt{2}\sin \alpha \cos \alpha +\sqrt{2}({\sin }^2\alpha -{\cos }^2\alpha )=0$

$\Rightarrow \sqrt{2}\sin 2\alpha -\sqrt{2}\cos 2\alpha =1$

$\Rightarrow \sin (2\alpha -\frac{{\displaystyle \pi }}{{\displaystyle 4}})=\sin \frac{\pi }{6}\Rightarrow 2\alpha -\frac{\pi }{4}=n\pi +{(-1)}^n\frac{\pi }{6}$

for $n=0\Rightarrow \alpha =\frac{5\pi }{24}$

(C)

We have, $11x+y+\lambda z=-5$

                   $2x+3y+5z=3$

                    $8x-19y-39z=\mu$

has infinitely many solutions

$\therefore$                $\Delta =$$\left|\begin{array}{ccc}11& 1& \lambda \\ 2& 3& 5\\ 8& -19& -39\end{array}\right|$$=0$

$\Rightarrow 11(-22)-1(-118)+\lambda (-62)=0$

$\Rightarrow -124-62\lambda =0$

$\Rightarrow \lambda =-2$

Also, ${\Delta }_3=$$\left|\begin{array}{ccc}11& 1& -5\\ 2& 3& 3\\ 8& -19& \mu \end{array}\right|$$=0$

$\Rightarrow 11(3\mu +57)-1(2\mu -24)-5(-38-24)$

$\Rightarrow 31\mu +961=0\Rightarrow \mu =-31$

So, ${\lambda }^4-\mu ={(-2)}^4+31=16+31=47$

(C)

Since the system of equations has infinitely many solutions

$\therefore$    $\Delta =0 \text{and} {\Delta }_3=0$

$\Rightarrow$$\left|\begin{array}{ccc}1& 1& 1\\ 2& 5& 5\\ 1& 2& m\end{array}\right|$$=0$ and $\left|\begin{array}{ccc}1& 1& 4\\ 2& 5& 17\\ 1& 2& n\end{array}\right|$$=0$

$\Rightarrow (5m-10)-(2m-5)+(-1)=0$ and $(5n-34)-(2n-17)+4(-1)=0$

$\Rightarrow 3m-6=0$ and $3n-21=0$

$\Rightarrow m=2$ and $n=7$

Clearly $m^2+n^2-mn=4+49-14=39$

(C)

We have, $x+4y-z=\lambda$

$7x+9y+\mu z=-3; 5x+y+2z=-1$

$\left[\begin{array}{ccc}1& 4& -1\\ 7& 9& \mu \\ 5& 1& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}\lambda \\ -3\\ -1\end{array}\right]$

$A=\left[\begin{array}{ccc}1& 4& -1\\ 7& 9& \mu \\ 5& 1& 2\end{array}\right], B=\left[\begin{array}{c}\lambda \\ -3\\ -1\end{array}\right]$

$AX=B\Rightarrow X=A^{-1}B=\frac{\text{adj }A}{\left|A\right|}B$

Since, the system has infinitely many solutions,

$\therefore$   $\left|A\right|=0 \text{and} \left(\text{adj }A\right)·B=0.$

Now, $\left|A\right|$$=0\Rightarrow 18-\mu - 4(14-5\mu )-1(7-45)= 0$

$\Rightarrow 18-\mu -56+20\mu +38=0$

$\Rightarrow 19\mu =0\Rightarrow \mu =0$

Also, $\text{adj }A=\left[\begin{array}{ccc}18& -9& 9\\ -14& 7& -7\\ -38& 19& -19\end{array}\right]$

         $(\text{adj }A·B)=0$

$\Rightarrow \left[\begin{array}{ccc}18& -9& 9\\ -14& 7& -7\\ -38& 19& -19\end{array}\right]\left[\begin{array}{c}\lambda \\ -3\\ -1\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

      $18\lambda +27-9=0\Rightarrow 18\lambda =-18\Rightarrow \lambda =-1$

Hence, $2\mu +3\lambda =2\left(0\right)+3(-1)=-3$

(C)

$3x+5y+\lambda z=3$

$7x+11y-9z=2$

$97x+155y-189z=\mu$

$\left[\begin{array}{ccc}3& 5& \lambda \\ 7& 11& -9\\ 97& 155& -189\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}3\\ 2\\ \mu \end{array}\right]$

$AX=B$

For infinitely many solutions,

$\left|A\right|$$=0 \text{and} \left(\text{adj }A\right)B=0$

$\Rightarrow$$\left|\begin{array}{ccc}3& 5& \lambda \\ 7& 11& -9\\ 97& 155& -189\end{array}\right|$$=0$

$\Rightarrow -2052+2250+18\lambda =0$

$\Rightarrow \lambda =-11$

$\text{adj }A=\left[\begin{array}{ccc}-684& -760& 76\\ 450& 500& -50\\ 18& 20& -2\end{array}\right]$

$\left(\text{adj }A\right)B=\left[\begin{array}{ccc}-684& -760& 76\\ 450& 500& -50\\ 18& 20& -2\end{array}\right]\left[\begin{array}{c}3\\ 2\\ \mu \end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

$\Rightarrow 54+40-2\mu =0$

$\Rightarrow 2\mu =94 \Rightarrow \mu =47$

Hence, $\mu +2\lambda =25$

(D)

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

where $A=\left[\begin{array}{ccc}2& 3& -1\\ 1& \alpha & 3\\ 3& -1& \beta \end{array}\right], B=\left[\begin{array}{c}5\\ -4\\ 7\end{array}\right]$ and $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$

Using Cramer's rule for infinite solutions,

$D=D_1=D_2=D_3=0$

$D_3=$$\left|\begin{array}{ccc}2& 3& 5\\ 1& \alpha & -4\\ 3& -1& 7\end{array}\right|$$=2(7\alpha -4)-3(7+12)+5(-1-3\alpha )=0$

$\Rightarrow \alpha =-70$

Similarly,

$D_2=$$\left|\begin{array}{ccc}2& 5& -1\\ 1& -4& 3\\ 3& 7& \beta \end{array}\right|$$=2(-4\beta -21)-5(\beta -9)-1(7+12)=0$

$\Rightarrow \beta =-\frac{16}{13}$

$\therefore$     $13\alpha \beta =13\times (-70)\times \left(\frac{-16}{13}\right)=1120$

(B)

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

Where $A=\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\\ 4& 3& \lambda \end{array}\right], X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],$and $B=\left[\begin{array}{c}5\\ 9\\ \mu \end{array}\right]$

For infinitely many solutions, $D=0\Rightarrow$$\left|\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\\ 4& 3& \lambda \end{array}\right|$$=0$

$\Rightarrow \lambda =-13$

Also, $D_1=0\Rightarrow$$\left|\begin{array}{ccc}5& 2& 3\\ 9& 3& 1\\ \mu & 3& -13\end{array}\right|$$=0\Rightarrow \mu =15$

So, $\lambda +2\mu =-13+30=17$

(A)

We have, $\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 2\lambda \\ 1& 3& 4{\lambda }^2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}4\mu \\ 10\mu \\ {\mu }^2+15\end{array}\right]$

For $\lambda =\frac{1}{2}$ and $\mu =15,$ we get $\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 1\\ 1& 3& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}60\\ 150\\ 240\end{array}\right]$

System is consistent but does not have unique solution as matrix have zero determinant because two columns are same.

Also, let $\lambda \ne \frac{1}{2}$ and $\lambda =\mu =1,$ then we have

$\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 2\\ 1& 3& 4\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}4\\ 10\\ 16\end{array}\right]$

$\Rightarrow x+y+z=4; x+2y+2z=10; x+3y+4z=16$

On solving these equations, we get $(-2,6,0)$ as unique solution.

Clearly, for $\lambda =\frac{1}{2}$ and $\mu \ne 1$ i.e., $\mu =15$ the system is consistent and have infinite solution.

So, statement (a) is not correct.

(A)

$\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& {\lambda }^2\\ 1& 3& \lambda \end{array}\right|$$=0\Rightarrow 2{\lambda }^2-\lambda -1=0$

$\Rightarrow \lambda =1,-\frac{1}{2} \text{and}$$\left|\begin{array}{ccc}1& 1& 5\\ 2& {\lambda }^2& 9\\ 3& \lambda & \mu \end{array}\right|$$=0\Rightarrow \mu =13$

For infinite solution, $\lambda =1$ and $\mu =13$

For unique solution $\lambda \ne 1$

For no solution,$\lambda =1$ and $\mu \ne 13$

If $\lambda \ne 1$ and $\mu \ne 13$

Considering the case when $\lambda =-\frac{1}{2}$ and $\mu \ne 13$ this will generate no solution case.

(C)

Given, $x-2y+z=-4$

$2x+\alpha y+3z=5, 3x-y+\beta z=3$

      $\Delta =$$\left|\begin{array}{ccc}1& -2& 1\\ 2& \alpha & 3\\ 3& -1& \beta \end{array}\right|$$=0$

$\Rightarrow \alpha \beta -18-2-3\alpha +3+4\beta =0=\alpha \beta -3\alpha +4\beta -17=0$

       $\Delta y=$$\left|\begin{array}{ccc}1& -4& 1\\ 2& 5& 3\\ 3& 3& \beta \end{array}\right|$$=0$

$\Delta y=1(5\beta -9)+4(2\beta -9)+1(6-15)$

        $=5\beta -9+8\beta -36+6-15$

$\Rightarrow 13\beta -54=0\Rightarrow 13\beta =54\Rightarrow \beta =\frac{54}{13}$

       $\Delta z=$$\left|\begin{array}{ccc}1& -2& -4\\ 2& \alpha & 5\\ 3& -1& 3\end{array}\right|$$=0$

       $\Delta z=1(3\alpha +5)+2(6-15)-4(-2-3\alpha )$

$\Rightarrow 3\alpha +5+12-30+8+12\alpha$

$\Rightarrow 15\alpha -5=0\Rightarrow 15\alpha =5\Rightarrow \alpha =\frac{5}{15}=\frac{1}{3}$

So,$12\alpha +13\beta =12\times \frac{1}{3}+54\Rightarrow 4+54=58$