Class 12th

From the given structure of CuSO₄ . 5H₂O Cu (II) ion and oxygen bonds are present but ligands coordinating with Cu (II) ion are not O and S both.

1st ionization energy of N>O in oxygen atom, 2 electrons out of 4 electron of 2p orbital resulting in an increased electron repulsion.

 $\overrightarrow{a}*\left(\overrightarrow{b}*\overrightarrow{c}\right)=3\overrightarrow{b}-\overrightarrow{c}=\overrightarrow{u}$

$\overrightarrow{b}*\left(\overrightarrow{c}*\overrightarrow{a}\right)=\overrightarrow{c}-2\overrightarrow{a}=\overrightarrow{v}$

$\overrightarrow{c}*\left(\overrightarrow{b}*\overrightarrow{a}\right)=3\overrightarrow{b}-2\overrightarrow{a}=\overrightarrow{w}$

$\therefore \overrightarrow{u}+\overrightarrow{v}=\overrightarrow{w}$

so vectors 

$\overrightarrow{u},\overrightarrow{v}and\overrightarrow{w}$

are coplanar, hence their Scalar triple product will be zero.

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

$?{\mathcal{l}}_{1}and{\mathcal{l}}_2$ are perpendicular, so

$3*1+\left(-2\right)\left(\frac{\alpha }{2}\right)+0*2=0$

a = 3

Now angle between ${\mathcal{l}}_{2}and{\mathcal{l}}_3$ ,

$cos\theta =\frac{1\left(-3\right)+\frac{\alpha }{2}\left(-2\right)+2\left(4\right)}{\sqrt[]{1+\frac{{\alpha }^2}{4}+4}.\sqrt[]{9+4+16}}$

$\Rightarrow cos\theta =\frac{\frac{2}{29}}{2}\Rightarrow \theta =co{s}^{-1}\left(\frac{4}{29}\right)=se{c}^{-1}\left(\frac{29}{4}\right)$

Let P (at², 2 at) where

$a=\frac{3}{2}$                

T : yt = x + at² so point Q is

$\left(-a,at-\frac{a}{t}\right)$                

N : y = -tx + 2at + at³ passes through (5, -8)

$-8=-5t+3t+\frac{3}{2}{t}^3$

⇒ $3t^3-4t+16=0$                

⇒ t = -2

So ordinate of point Q is $-\frac{9}{4}$  

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

Equation of L₁ = is

$\frac{xsec\theta }{4}-\frac{ytan\theta }{2}=1$ ….(i)

Equation of line L₂ is

$\frac{xtan\theta }{2}+\frac{ysec\theta }{4}=0$ ….(ii)

$?$ Required point of intersection of L₁ and L₂ is (x₁, y₁) then

$\frac{x_{1}sec\theta }{4}-\frac{y_{1}tan\theta }{2}-1=0$ ….(iii)

$and\frac{y_{1}sec\theta }{4}+\frac{x_{1}tan\theta }{2}=0$ ……(iv)

From equations (iii) and (iv)

$sec\theta =\frac{4x_1}{x_1^2+y_1^2}andtan\theta =\frac{-2y_1}{x_1^2+y_1^2}$        

$\therefore$ Required locus of (x₁, y₁) is

${\left(x^2+y^2\right)}^2=16x^2-4y^2$   

$\begin{array}{l}\therefore \alpha =16,\beta =-4\\ \therefore \alpha =\beta =12\end{array}$     

Equation of perpendicular bisector of AB is

$y-\frac{3}{2}=-\frac{1}{5}\left(x-\frac{5}{2}\right)\Rightarrow x+5y=10$

Solving it with equation of given circle,

${\left(x-5\right)}^2+{\left(\frac{10-x}{5}-1\right)}^2=\frac{13}{2}$

$\Rightarrow x-5=\pm \frac{5}{2}\Rightarrow x=\frac{5}{2}or\frac{15}{2}$

But

$x\ne \frac{5}{2}$

because AB is not the diameter.

So, centre will be

$\left(\frac{15}{2},\frac{1}{2}\right)$

Now,

$r^2={\left(\frac{15}{2}-2\right)}^2+{\left(\frac{1}{2}+1\right)}^2=\frac{65}{2}$