# 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

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.
$\Rightarrow\:(a,b,c)=(5,2,2)$
$\therefore$ The equation of the plane becomes $5x+2y+2z+d=0$
But given that the plane passes through $(-1,2,1)$
$\therefore\:-5+4+2+d=0$
$\Rightarrow\:d=-1$
$\Rightarrow\:$ The equation of the plane is $5x+2y+2z-1=0$