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If the line $\large\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z-2}{2}$ lie in the plane $x+3y-\alpha z+\beta=0, \;\;then\:\:(\alpha,\beta)=?$

$\begin {array} {1 1} (A)\;(-6, -17) & \quad (B)\;(-5, 5) \\ (C)\;(6, -17) & \quad (D)\;(5, -15) \end {array}$

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1 Answer

Given equation of the line is $\large\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z-2}{2}$
Equation of the plane: $x+3y-\alpha z+\beta=0$
$\Rightarrow$ A known point on the line is $ (2,1,2)$
D.R. of the line is $(3,-5,2)$ and D.R. of normal to the plane is $(1,3,-\alpha)$
Since the line lie on the plane every point on the line lie on the plane.
$\therefore \:(2,1,2)$ satis fy the plane eqn.
Also the normal to the plane is $\perp$ to every line on the plane.
$\therefore\:(3,-5,2).(1,3,-\alpha)=0$ $\Rightarrow\:3-15-2\alpha=0.\:or\:\:\alpha=-6............(ii)$
and $\beta=-17$
answered Dec 23, 2013 by rvidyagovindarajan_1

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