Probability

$a=\left|\frac{4.2-3\left(-1\right)-21}{5}\right|$

$=\left|\frac{8+3-21}{5}\right|=2$

$\therefore L=4a=8$

Consider the equation of plane,

$P:\left(2x+3y+z+20\right)+\lambda \left(x-3y+5z-8\right)=0$

$?$ Plane P is perpendicular to 2x + 3y + z + 20 = 0

So,  $4+2\lambda +9-9\lambda +1+5\lambda =0$

0 $\lambda =7$

P : 9x – 18y + 36z – 36 = 0

Or P : x – 2y + 4z = 4

If image of

$\left(2,-\frac{1}{2},2\right)$

In plane P is (a, b, c) then

$\frac{a-2}{1}=\frac{b+\frac{1}{2}}{-2}=\frac{c-2}{4}$

and $\left(\frac{a+2}{2}\right)-2\left(\frac{b-\frac{1}{2}}{2}\right)+4\left(\frac{c+2}{2}\right)=4$

clearly

$a=\frac{4}{3},b=\frac{5}{6}andc=-\frac{2}{3}$

So, a : b : c = 8 : 5 : 4

Total number of numbers from given condition = n (s) = 26

Every required number is of the form

${}^{}$ $A=7.\left(10^{a_1}+10^{a_2}+10^{a_3}+....\right)$  + 111111

Here 111111 is always divisible by 21.

$$  $\therefore$ Required probability = $\frac{22}{2^5}=p$ $\frac{}{{}^{}}$ $\frac{}{}$ $\therefore \frac{11}{32}$ = p $$ $\therefore$ 96p = 33

The given two lines are coplanar

$\therefore \left|\begin{array}{ccc}\hfill 0\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill -3\hfill \\ \hfill 1-\alpha \hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|=0\Rightarrow \alpha =\frac{5}{3}$

Now, $\overrightarrow{n}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill -3\hfill \end{array}\right|=\widehat{i}\left(-9\right)-\widehat{j}\left(2\right)+\widehat{k}\left(-6\right)=\left(9,2,6\right)$

Equation of plane :

$=\left|\frac{\left(9.\frac{5}{3}+0+0-13\right)}{\sqrt[]{81+36+4}}\right|=\frac{2}{\sqrt[]{121}}=\frac{2}{11}$

We have,  $1-$  (probability of all shots result in failure)  $>\frac{1}{4}$

$\begin{array}{rr}& \Rightarrow 1-{\left(\frac{9}{10}\right)}^{\mathrm{n}}>\frac{1}{4}\\ & \Rightarrow \frac{3}{4}>{\left(\frac{9}{10}\right)}^{\mathrm{n}}\Rightarrow \mathrm{n}\ge 3\end{array}$

$A^2=\left(\begin{array}{ll}\mathrm{c}\mathrm{o}\mathrm{s}?2\theta & i\mathrm{s}\mathrm{i}\mathrm{n}?2\theta \\ i\mathrm{s}\mathrm{i}\mathrm{n}?2\theta & \mathrm{c}\mathrm{o}\mathrm{s}?2\theta \end{array}\right)$

Similarly, $A^5=\left(\begin{array}{cc}\mathrm{c}\mathrm{o}\mathrm{s}?5\theta & i\mathrm{s}\mathrm{i}\mathrm{n}?5\theta \\ i\mathrm{s}\mathrm{i}\mathrm{n}?5\theta & \mathrm{c}\mathrm{o}\mathrm{s}?5\theta \end{array}\right)=\left(\begin{array}{ll}a& b\\ c& d\end{array}\right)$

(1) $a^2+b^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta -{\mathrm{s}\mathrm{i}\mathrm{n}}^2?5\theta =\mathrm{c}\mathrm{o}\mathrm{s}?10\theta =\mathrm{c}\mathrm{o}\mathrm{s}?{75}^{?}$

(2) $a^2-d^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta -{\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta =0$

(3) $a^2-b^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta +{\mathrm{s}\mathrm{i}\mathrm{n}}^2?5\theta =1$

(4) $a^2-c^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta +{\mathrm{s}\mathrm{i}\mathrm{n}}^2?5\theta =1$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}WeknowthatE\left(X\right)={\displaystyle \sum _{i=1}^n{X}_i{P}_i}\\ =\left(-4\right)\left(0.1\right)+\left(-3\right)\left(0.2\right)+\left(-2\right)\left(0.3\right)+\left(-1\right)\left(0.2\right)+0\left(0.2\right)\\ =-0.4-0.6-0.6-0.2=-1.8\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat:Bagcontains5redand3blueballs.\\ ofgettingexactlyoneredballif3ballsarerandomlydrawnwithoutreplacement.\\ P\left(R\right).P\left(B\right).P\left(B\right)+P\left(B\right).P\left(R\right).P\left(B\right)+P\left(B\right).P\left(B\right).P\left(R\right)\\ =\frac{5}{8}.\frac{3}{7}.\frac{2}{6}+\frac{3}{8}.\frac{5}{7}.\frac{2}{6}+\frac{3}{8}.\frac{2}{7}.\frac{5}{6}=\frac{30}{336}+\frac{30}{336}+\frac{30}{336}=\frac{90}{336}=\frac{15}{56}\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Sol: