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If the line $\frac{x-2}{3}=\frac{y+1}{5}=\frac{z-2}{2}$ lies in the plane $\alpha x+2y+z-\beta =0$ , then $\left ( \alpha ,\beta \right )$ equals
Option: 1
$(4,8)$

Option: 2
$(-8,4)$

Option: 3
$(-4,-8)$

Option: 4
$(8,-4)$

Answers (1)

best_answer

Line $\frac{x-2}{3}=\frac{y+1}{5}=\frac{z-2}{2}$ lies in the plane $\alpha x+2y+z-\beta =0$

$\because$ So, point $\left ( 2,-1,2 \right )$ lies on the plane

So, $\alpha \left ( 2 \right )+2\left ( -1 \right )+2-\beta =0$

$2\alpha -\beta =0$

and direction ratio of line is perpendicular to the normal vector of plane.

$\Rightarrow \left ( 3 \right )\left ( \alpha \right )+5\left ( 2 \right )+2\left ( 1 \right )=0$

$\alpha =-4$

$\beta =-8$

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