# find equation of plane containing (1,2,3) and perpendicular to the place 3x- y +2z= 3 and parallel to the line x-1/3=y/2=z/-2.write equation in dot product form

Let the equation of the plane be $ax+by+cz+d=0$
$\Rightarrow\:$ D.R. of the normal to the plane is $(a,b,c)$
Given: This plane is $\perp$ to the plane $3x-y+2z-3=0$
$\therefore\: (a,b,c).(3,-1.2)=0$ $\Rightarrow\:3a-b+2c=0.....(i)$
Also given that the plane is parallel to the line $\large\frac{x-1}{3}=\frac{y}{2}=\frac{z}{-2}=\lambda\:\: (say)$
$\Rightarrow\:$ Normal to the plane is $\perp$ to the line.
$\Rightarrow\:(a,b,c).(3,2,-2)=0\:\:\Rightarrow\:3a+2b-2c=0.......(ii)$
It is given that the plane passes through the point $(1,2,3)$.
Solving $(i)\:\:and\:\:(ii)$ we get
$\large\frac{a}{2-4}=\frac{-b}{-6-6}=\frac{c}{6+3}$
$\Rightarrow\:a=-2,\:b=12, \:c=9$
Substituting the values in the equation of the plane we get
$-2x+12y+9z+d=0$
But given that the point $(1,2,3)$ lies on the plane.
$\therefore\:-2+24+27+d=0$
$\Rightarrow\:d=-49$
$\therefore$ The required equation of the plane is
$2x-12y-9z+49=0$
In dot product form the equation is :   $\overrightarrow r.(2\hat i-12\hat j-9\hat k)+49=0$

edited Nov 13, 2013