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Let the plane P pass through the intersection of the planes $2 x+3 y-z=2$ and $x+2 y+3z=6$ and be perpendicular to the plane $2x + y - z = 0$ . If d is the distance of P form the point (–7,1,1,) then $d^{2}$ is equal to :
Option: 1
$\frac{250}{83}$

Option: 2
$\frac{250}{82}$

Option: 3
$\frac{15}{53}$

Option: 4
$\frac{25}{83}$

Answers (1)

best_answer

Plane P, is passing through intersection of the two planes, so,

$\begin{aligned} & 2 x+3 y-z-2+\lambda(x+2 y+3 z-6)=0 \\ & x(2+\lambda)+y(3+2 \lambda)+z(3 \lambda-1)-2-6 \lambda=0 \end{aligned}$

It is perpendicular with plane, $2 x+y-2+1=0$

So, $(2+\lambda) 2+(3+2 \lambda) 1+(3 \lambda-1)(-1)=0$

$\lambda=-8$

So, plane $p_1-6 x-13 y-25 z+46=0$

distance of plane p from the point $(-7,1,1)$

$$$ \begin{aligned} & d=\frac{|+42-13-25+46|}{\sqrt{36+169+625}}=\frac{50}{\sqrt{30}}= \\ & d^2=\frac{2500}{830}=\frac{250}{83} \end{aligned} $$ Ans. (1)$

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