Three Dimensional Geometry

  $\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}=r_1$

$A\left(3r_1+7,1-r_1,-2+r_1\right)$

and   $\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}=r_2$  

  $B\left(2r_2,7+3r_2,r_2\right)$

$A/q,\frac{3r_1-2r_2+7}{1}=\frac{3r_2+r_1+6}{4}=\frac{r_1-r_2-2}{2}$             

  $\Rightarrow r_1=-5,r_2=-3$

$\therefore A\left(-8,6,-7\right)andB\left(-6,-2,-3\right)$

AB² = 84

  $f\left(x\right)=\{\begin{array}{l}\left|2x^2-3x-7\right|,x\le -1\\ \left[4x^2-1\right],-1<x<1\\ \left|x+1\right|+\left|x-2\right|,x\ge 1\end{array}$              

f(-1) = 1

f(1) = 3

Hence f(x) will be discontinuous at x = 1 and also 4x² – 1 = 0 , 1 , 2

$\Rightarrow x=\pm \frac{1}{2},\pm \frac{1}{\sqrt[]{2}},\pm \frac{\sqrt[]{3}}{2}$          

Let P (x, y, z) be any point on plane P₁ then

${\left(x+4\right)}^2+{\left(y-2\right)}^2+{\left(z-1\right)}^2={\left(x-2\right)}^2+{\left(y+2\right)}^2+{\left(z-3\right)}^2$

$\therefore cos\theta =\frac{\left|6-2+3\right|}{14}\Rightarrow \theta =\frac{\pi }{3}$

S. D. = $\frac{1}{\sqrt[]{3}}$

$b_1*b_2=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill \lambda \hfill \\ \hfill 1\hfill & \hfill 4\hfill & \hfill 5\hfill \end{array}\right|=\widehat{i}\left(15-4\lambda \right)+\widehat{j}\left(\lambda -10\right)+\widehat{k}\left(5\right)$

$\frac{\left|\left(-\widehat{i}-2\widehat{j}-2\widehat{k}\right).\left\{\left(15-4\lambda \right)\widehat{i}+\left(\lambda -10\right)\widehat{j}+5\widehat{k}\right\}\right|}{\sqrt[]{{\left(15-4\lambda \right)}^2+{\left(\lambda -10\right)}^2+25}}=\frac{1}{\sqrt[]{3}}$

$\Rightarrow \lambda =16$

 $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=\frac{-2\left(1+4+2-16\right)}{1+2^2+2^2}$

(x, y, z) = (3, 6, 5)

now point Q and line both lies in the plane.

So, equation of plane is

$\left|\begin{array}{ccc}\hfill x\hfill & \hfill y\hfill & \hfill z+1\hfill \\ \hfill 3\hfill & \hfill 6\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right|=0$

2x – z = 1

option (B) satisfies.

Let AB $\equiv x-2y+1=0$

AC $\equiv 2x-y+1=0$

So vertex A = (1, 1)

altitude from B is perpendicular to AC and passing through orthocentre.

So, BH = x + 2y – 7 = 0

CH = 2x + y – 7 = 0

now solve AB & BH to get B (3, 2) similarly CH and AC to get C (2, 3) so centroid is at (2, 2)

Line $\perp$ to the normal

⇒ 3p + 2q – 1 = 0

$\left(2,-1,-3\right)$ lies in the plane 2p + q = 8

From here p = 15, q = -22

Equation of plane 15x – 22y + z – 5 = 0

Distance from origin = $\left|\frac{5}{\sqrt[]{15^2+{\left(-22\right)}^2+1^2}}\right|$$=$√5/142

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,

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

  $E:\frac{x^2}{4}+\frac{y^2}{2}=1$

any pt on it is P  $\left(2cos\theta ,\sqrt[]{2}sin\theta \right)$

$\therefore$ M (h, k) be mid point of P & A (4, 3)

$\Rightarrow {\left(h-2\right)}^2+{\left(\frac{2k-3}{\sqrt[]{2}}\right)}^2=1$        

Required locus (x – 2)² +   $\frac{{\left(y-\frac{3}{2}\right)}^2}{\frac{1}{2}}=1$

$\therefore e=\sqrt[]{1-\frac{1}{2}}=\frac{1}{\sqrt[]{2}}$

Let PT perpendicular to QR

$\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}=\lambda \Rightarrow T\left(2\lambda -1,3\lambda -2,2\lambda +1\right)$ therefore

$2\left(2\lambda -5\right)+3\left(3\lambda -4\right)+2\left(2\lambda -6\right)=0\Rightarrow \lambda =2$

$T\left(3,4,5\right)\therefore PT=\sqrt[]{1+4+4}=3\therefore QT=\sqrt[]{26-9}=\sqrt[]{17}$

$\therefore \Delta PQR=\frac{1}{2}*2\sqrt[]{17}*3=3\sqrt[]{17}$

Therefore square of $ar\left(\Delta PQR\right)$ = 153.