Maths

$9*9*3=81*3=243$

Clearly all equations can be $\lambda x-2y-6z=0$  form where $\lambda =1,\frac{4}{3},\frac{6}{5}$  

So for $x=0y=-3z$

So $x^2+y^2+z^2=10z^2$  (as  ) $x=0\mathrm{}y=-3z$

ATQ $10\le 10z^2\le 200$

$1\le z^2\le 20\Rightarrow z=\pm 1,\pm 2,\pm 3,\pm 4$

By property

If first and Last term of Andhra Pradesh and HP are equal (of nn  terms) then $a_r{H}_S=$  first *  x last term

If $r+s=n+1$ r+s=n+1

Equation of tangent of at $y=\mathrm{l}\mathrm{o}\mathrm{g}?x$  at $\left(e^{\lambda },\lambda \right)$  is $y-\lambda =\frac{1}{e^{\lambda }}\left(x-e^{\lambda }\right)$

It cuts y   axis at $(0,\lambda ,-1)$

Also normal to $y^2=8x$    at $\left(\mathrm{7,4}\right)$    is $y-4=-(x-2)$  

  $y=-x+6\mathrm{}$  Cuts y axis at $\left(\mathrm{0,6}\right)\mathrm{}$  SO ATQ $6=-1+\lambda \Rightarrow \lambda =7$

Plane containing both line $x=1$

Image of point $(\mathrm{2,1},3)$  in the plane $\frac{\alpha -2}{1}=\frac{\beta -1}{0}=\frac{\gamma -3}{0}=-2\left(1\right)$

$\alpha =0,\beta =1,\gamma =3;\alpha +\beta +\gamma =4$

Single digit factor of $12\to \mathrm{1,2},\mathrm{3,4},6$

  $(x,y,z)$  where , $xyz=12$

$(\mathrm{6,2},1)(\mathrm{4,3},1)(\mathrm{3,2},2)$

$\begin{array}{ccc}\downarrow & \downarrow & \downarrow \\ \text{6Ways}& \text{6 Ways}& \text{3Ways}\end{array}$

 Total 15 Ways

Equation of normal   $y=mx-2m-m^3$ must pass through centre $\left(\mathrm{0,3}\right)$

$\begin{array}{rr}& m^3+2m+3=0\\ & m^2(m+1)-m(m+1)+3\cdot 1(m+1)=0\\ & (m+1)\left(m^2-m+3\right)=0\\ & m=-1\end{array}$

  $\therefore \mathrm{}$ point of contact of normal at parabola is $\left({\mathrm{a}\mathrm{m}}^2,-2\mathrm{a}\mathrm{m}\right)=\left(\mathrm{1,2}\right)$

So distance between parabola and circle is $=\sqrt[]{2}-1$

  $x-cy-bz=0$

$cx-y+az=0$

$bx+ay-z=0$

Equation of any plane passing through the line of intersection of planes (1) and (2)

$x-cy-bz+\lambda (cx-y+az)=0$

$(1+\lambda c)x-(c+\lambda )y+(-b+a\lambda )z=0$

If (3) &  (4) is same plane.

$\frac{1+c\lambda }{b}=\frac{-(c+\lambda )}{a}=\frac{-b+a\lambda }{-1}$

(i)

(ii) (iii)

By (i)  & (ii) $\lambda =-\frac{(a+bc)}{ac+b}$

By (ii) & (iii)

$\lambda =\frac{-(ab+c)}{1-a^2};a^2+b^2+c^2+2abc=1$

$n(A\cup B\cup C)+n(A\cup B\cup C{)}^{\mathrm{\text{'}}}=50$

$n(\cup )=50$

$n(A\cup B\cup C{)}^{\mathrm{\text{'}}}=n(\cup )-n(A\cup B\cup C)\Rightarrow 50-29\Rightarrow 21$

$n(A\cup B\cup C)=n\left(A\right)+n\left(B\right)+n\left(C\right)-n(A\cap B)-n(B\cap C)-n(C\cap A)-n(A\cap B\cap C)$

$\Rightarrow \mathrm{}10+15+20-5-6-10-5\Rightarrow 29$

$n(A\cup B\cup C{)}^{\mathrm{\text{'}}}=21$

$E\left(\frac{-a{x}_1+b{x}_2+c{x}_3}{-a+b+c},\frac{-a{y}_1+b{y}_2+c{y}_3}{-a+b+c}\right)E\left(\frac{-4*0+4*3+0*5}{-4+3+5},\frac{-4*3+3*0+5*0}{-4+3+5}\right)$