Question

Find the equation of the plane passes through the point $(-1,3,2)$ and is perpendicular to planes $x+2y+2z=5$ and $3x+3y+2z=8$.

Answer 1

The equation of the plane passes through the point $(-1,3,2)$ is

$a(x+1)+b(y-3)+c(z-2)=0$ ........(1)

If this plane is perpendicular to the planes.

$x+2y+2z=5$

and $3x+3y+2z=8$

then $a+2b+2c=0$ ..........(2)

and $3a+3b+2c=0$ .........(3)

$\Rightarrow \frac{a}{4-6}=\frac{b}{6-2}=\frac{c}{3-6}$

$\Rightarrow \frac{a}{-2}=\frac{b}{+4}=\frac{c}{-3}$

$\Rightarrow \frac{a}{2}=\frac{b}{-4}=\frac{c}{3}=K$

Let
$\therefore a=2K,b=-4K,c=3K$

Putting the values of $a,b,c$ in equation (1), we get the required equation of plane.

$2K(x+1)-4K(y-3)+3K(z-2)=0$

$\Rightarrow 2(x+1)-4(y-3)+3(z-2)=0$

$\Rightarrow 2x+2-4y+12+3z-6=0$

$\Rightarrow 2x-4y+3z+8=0$.