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find the equation of plane through (-1,2,1) and perpendicular to the line joining (-3,1,2) and (2,3,4) find the distance of plane from origin

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Let the equation of the plane be $ax+by+cz+d=0$
$\Rightarrow\: $ D.R. of normal to the plane is $(a,b,c)$
Given that the plane is $\perp$ to the line joining $A(-3,1,2)\:\:and\:\:B(2,3,4)$
$\therefore\: \overline {AB}=(5,2,2)$ becomes normal to the plane.
$\therefore$ The equation of the plane becomes $ 5x+2y+2z+d=0$
But given that the plane passes through $(-1,2,1)$
$\Rightarrow\:$ The equation of the plane is $5x+2y+2z-1=0$
answered Nov 13, 2013 by rvidyagovindarajan_1

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