(3)

Given : x + 2y – 3z = 2

$2x+\lambda y+5z=5$

$14x+3y+\mu z=33$

For infinitely many solutions, we have $∆=0$

$\Rightarrow$$\left|\begin{array}{ccc}1& 2& –3\\ 2& \lambda & 5\\ 14& 3& \mu \end{array}\right|$$=0$

$\Rightarrow 1(\lambda \mu –15)–2(2\mu –70)–3(6–14\lambda )=0$

$\Rightarrow \lambda \mu +42\lambda –4\mu +107=0$

We also have, ${△}_1=2\lambda \mu +99\lambda –10\mu +225$

${△}_2=\mu –13, {△}_3=5\lambda +5$

Now, ${△}_2=0 \mathrm{and} {△}_3=0$

$\therefore 13–\mu =0 \mathrm{and} 5\lambda +5=0 \Rightarrow \mu =13 \Rightarrow \lambda =–1$

Thus, we get $\lambda +\mu =13+(–1)=13–1=12$

(4)

We have, x + y + z = 1

x + 2y +4z = m and x + 4y + 10z = $m^2$

$\therefore △=$$\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 4\\ 1& 4& 10\end{array}\right|$$=1\left(4\right)–1\left(6\right)+1\left(2\right)=0$

For infinite many solutions, $△x=△y=△z=0$

Now, $△x=0 \Rightarrow$$\left|\begin{array}{ccc}1& 1& 1\\ m& 2& 4\\ m^2& 4& 10\end{array}\right|$$=0$

$\Rightarrow 1\left(4\right)–1(10m–4m^2)+1(4m–2m^2)=0$

$\Rightarrow 4–10m+4m^2+4m–2m^2=0$

$\Rightarrow 2m^2–6m+4=0$

$\Rightarrow m^2–3m+2=0 \Rightarrow m=1,2$

$\therefore \alpha =1, \beta =2$

Now, $\sum _{n=1}^{10}(n^{\alpha }+n^{\beta })=\sum _{n=1}^{10}n+\sum _{n=1}^{10}{n}^2$

$=\frac{10\left(11\right)}{2}+\frac{10\left(11\right)\left(21\right)}{6}$ = 55 + 385 = 440.

(4)

Given system of equations are

$x+y+az=b$

$2x+5y+2z=6$

$x+2y+3z=3$

For infinitely many solutions, $\Delta =0, {\Delta }_1=0, {\Delta }_2=0, {\Delta }_3=0$

$\Delta =$$\left|\begin{array}{ccc}1& 1& a\\ 2& 5& 2\\ 1& 2& 3\end{array}\right|$$=0\Rightarrow 1(15-4)-1(6-2)+a(4-5)=0$

$\Rightarrow 11-4-a=0\Rightarrow a=7$

${\Delta }_1=$$\left|\begin{array}{ccc}b& 1& a\\ 6& 5& 2\\ 3& 2& 3\end{array}\right|$$=0$

$\Rightarrow b(15-4)-1(18-6)+a(12-15)=0$

$\Rightarrow 11b-12+7(-3)=0\Rightarrow b=\frac{33}{11}=3$

$\therefore$  $2a+3b=2\left(7\right)+3\left(3\right)=23$

(3)

Let $\Delta =$$\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& \alpha \\ 1& 3& 5\end{array}\right|$$=1(10-3\alpha )-1(5-\alpha )+1$$=6-2\alpha$

For a unique solution, $\Delta \ne 0\Rightarrow \alpha \ne 3$

Now, if $\alpha =3$, then

${\Delta }_1=$$\left|\begin{array}{ccc}6& 1& 1\\ 10& 2& 3\\ \beta & 3& 5\end{array}\right|$$=6(10-9)-1(50-3\beta )+1(30-2\beta )$$=\beta -14$

${\Delta }_2=$$\left|\begin{array}{ccc}1& 6& 1\\ 1& 10& 3\\ 1& \beta & 5\end{array}\right|$$=1(50-3\beta )-6(5-3)+1(\beta -10)$ $=28-2\beta =-2(\beta -14)$

${\Delta }_3=$$\left|\begin{array}{ccc}1& 1& 6\\ 1& 2& 10\\ 1& 3& \beta \end{array}\right|$$=1(2\beta -30)-1(\beta -10)+6(3-2)$$=\beta -14$

Clearly, at $\beta =14$, ${\Delta }_i=0 \text{∀ i}=1,2,3$

$\therefore$   $\text{At }\alpha =3, \beta =14,\text{ the system of equations has infinitely many solutions.}$

(4)

For non-trivial solution, we have  $\left|\begin{array}{ccc}1& 1& \sqrt{3}\\ -1& \tan \theta & \sqrt{7}\\ 1& 1& \tan \theta \end{array}\right|$$=0$

$\Rightarrow 1({\tan }^2\theta -\sqrt{7})-1(-\tan \theta -\sqrt{7})+\sqrt{3}(-1-\tan \theta )=0$

$\Rightarrow {\tan }^2\theta -\sqrt{7}+\tan \theta +\sqrt{7}-\sqrt{3}-\sqrt{3}\tan \theta =0$

$\Rightarrow {\tan }^2\theta +\tan \theta -\sqrt{3}\tan \theta -\sqrt{3}=0$

$\Rightarrow \tan \theta (\tan \theta +1)-\sqrt{3}(\tan \theta +1)=0$

$\Rightarrow \tan \theta =-1\text{ or }\tan \theta =\sqrt{3}\Rightarrow \theta =\frac{-\pi }{4},\frac{3\pi }{4},\frac{\pi }{3},\frac{-2\pi }{3}$

So, $\frac{120}{\pi }\sum _{\theta \in S}\theta =\frac{120}{\pi }[\frac{{\displaystyle -\pi }}{{\displaystyle 4}}+\frac{{\displaystyle 3\pi }}{{\displaystyle 4}}+\frac{{\displaystyle \pi }}{{\displaystyle 3}}-\frac{{\displaystyle 2\pi }}{{\displaystyle 3}}]=\frac{120}{\pi }\times \frac{\pi }{6}=20$

(1)

We have,

$2x-y+3z=5$

$3x+2y-z=7$

$4x+5y+\alpha z=\beta$

By Cramer's rule,

$\Delta =$$\left|\begin{array}{ccc}2& -1& 3\\ 3& 2& -1\\ 4& 5& \alpha \end{array}\right|$$=7\alpha +35$

${\Delta }_1=$$\left|\begin{array}{ccc}5& -1& 3\\ 7& 2& -1\\ \beta & 5& \alpha \end{array}\right|$$=17\alpha -5\beta +130$

${\Delta }_2=$$\left|\begin{array}{ccc}2& 5& 3\\ 3& 7& -1\\ 4& \beta & \alpha \end{array}\right|$$=11\beta -\alpha -104$

${\Delta }_3=$$\left|\begin{array}{ccc}2& -1& 5\\ 3& 2& 7\\ 4& 5& \beta \end{array}\right|$$=7(\beta -9)$

For infinite many solutions, $\Delta ={\Delta }_1={\Delta }_2={\Delta }_3=0$

$\Rightarrow \alpha =-5 \text{and} \beta =9$

So, option (a) is incorrect and option (b) is correct.

For unique solution, $\Delta \ne 0\Rightarrow \alpha \ne -5$ and $\beta$ can be any value.

$\therefore \text{Option (c) is correct.}$

At $\alpha =-5$ and $\beta =8$

$\Delta =0 \text{and} {\Delta }_1=5\ne 0$                $\text{∴ Solution is inconsistent.}$

$\text{ So, option (4) is correct.}$

(1)

For infinitely many solutions, $\Delta ={\Delta }_1={\Delta }_2={\Delta }_3=0$

$\Delta =$$\left|\begin{array}{ccc}7& 11& \alpha \\ 5& 4& 7\\ 175& 194& 57\end{array}\right|$

$=7(228-1358)-11(285-1225)+\alpha (970-700)$

$=-7910+10340+270\alpha =0 \Rightarrow -2430+270\alpha =0 \Rightarrow \alpha =-9$

${\Delta }_1=$$\left|\begin{array}{ccc}13& 11& -9\\ \beta & 4& 7\\ 361& 194& 57\end{array}\right|$$=13(228-1358)-11(57\beta -2527)-9(194\beta -1444)$

$=-14690-627\beta +27797-1746\beta +12996 =26103-2373\beta =0$

$\Rightarrow \beta =11$           $\therefore \alpha +\beta +2=-9+11+2=4$

(2)

We have, system of linear equations

$2x+4y+2az=b$

$x+2y+3z=4$

$2x-5y+2z=8$

It can be written as $AX=B$,

where, $A=\left[\begin{array}{ccc}2& 4& 2a\\ 1& 2& 3\\ 2& -5& 2\end{array}\right], B=\left[\begin{array}{c}b\\ 4\\ 8\end{array}\right], X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$

$\left|A\right|$$=2(4+15)-4(2-6)+2a(-5-4)=54-18a$

If $\left|A\right|$$=0$, then $a=3$ and the system has infinitely many solutions.

If $\left|A\right|$$\ne 0$, then $a\ne 3$ and the system has a unique solution.

$A_{11}=19, A_{12}=4, A_{13}=-9, A_{21}=-38, A_{22}=-8, A_{23}=18, A_{31}=0, A_{32}=0, A_{33}=0$

$\text{adj}A=\left[\begin{array}{ccc}{A}_{11}& A_{21}& A_{31}\\ A_{12}& A_{22}& A_{32}\\ A_{13}& A_{23}& A_{33}\end{array}\right]=\left[\begin{array}{ccc}19& -38& 0\\ 4& -8& 0\\ -9& 18& 0\end{array}\right]$

For infinitely many solutions:

$\left(\text{adj}A\right)B=O\Rightarrow \left[\begin{array}{ccc}19& -38& 0\\ 4& -8& 0\\ -9& 18& 0\end{array}\right]\left[\begin{array}{c}b\\ 4\\ 8\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

$\Rightarrow 19b-152=0 \Rightarrow b=8$

(2)

$2x+y-z=5$

$2x-5y+\lambda z=\mu$

$x+2y-5z=7$

The system of equations has infinitely many solutions when $\Delta ={\Delta }_1={\Delta }_2={\Delta }_3=0$

$\Delta =0$

$\Rightarrow$$\left|\begin{array}{ccc}2& 1& -1\\ 2& -5& \lambda \\ 1& 2& -5\end{array}\right|$$=0$

$\Rightarrow 2(25-2\lambda )-1(-10-\lambda )-1(4+5)=0$

$\Rightarrow 50-4\lambda +10+\lambda -9=0 \Rightarrow -3\lambda =-51\Rightarrow \lambda =17$

${\Delta }_3=0$

$\left|\begin{array}{ccc}2& 1& 5\\ 2& -5& \mu \\ 1& 2& 7\end{array}\right|$$=0$$\Rightarrow 2(-35-2\mu )-1(14-\mu )+5(4+5)=0$

$\Rightarrow -70-4\mu -14+\mu +45=0\Rightarrow -3\mu =39\Rightarrow \mu =-13$

$\therefore$   ${(\lambda +\mu )}^2+{(\lambda -\mu )}^2={(17-13)}^2+{(17+13)}^2=4^2+{30}^2=16+900=916$

(2)

$\text{We have,} -x+2y-9z=7 \text{... (i)}$

$-x+3y+7z=9 \text{... (ii)}$

$-2x+y+5z=8 \text{... (iii)}$

$-3x+y+13z=\lambda \text{... (iv)}$

From (i), we have $x=2y-9z-7 \text{... (v)}$

Substituting the value of $x$ in (ii) and (iii), we get

$y+16z=2$ and $-3y+23z=-6$

Solving these two equations, we get $y=2,z=0$.

$\Rightarrow x=2\times 2-9\times 0-7=4-7=-3 \text{[Using (v)]}$

Now, substituting the value of $x,y,z$ in equation (iv), we get

$-3\times (-3)+2+13\times \left(0\right)=\lambda \Rightarrow \lambda =11$

$\text{Distance of }(-3,2,0)\text{ from the plane }2x-2y+z=11\text{ is}$

$d=$$\left|\frac{{\displaystyle -6-4-11}}{{\displaystyle \sqrt{4+4+1}}}\right|$$=\frac{21}{3}=\text{7 units.}$