Question

Find the equation to the plane through the point $(-1,3,2)$ and perpendicular to the planes $x+2y+2z=11$ and $3x+3y+2z=15$.

Answer 1

Equation of plane passing through point $(-1,3,2)$
$a(x+1)+b(y-3)+c(z-2)=\mathrm{0....}\left(1\right)$
$x+2y+2z=\mathrm{11........}\left(2\right)$
and $3x++2z=\mathrm{15...}\left(3\right)$
$\because$ plane (1) is perpendicular to plane (2) and (3)
$\therefore$ by condition or perpendicularity
$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$
$a+2b+2c=\mathrm{0......}\left(4\right)$
$3a+3b+2c=\mathrm{0.....}\left(5\right)$
by solving the equation (4) and (5)
$\frac{a}{4-6}=\frac{b}{6-2}=\frac{c}{3-6}$
$\frac{a}{2}=\frac{b}{-4}=\frac{c}{3}$
Let $\frac{a}{2}=\frac{b}{-4}=\frac{c}{3}=K$
$\therefore$ $a=2K,b=-4K,c=3K$
Putting the value of a$a,b,c$ in equation (1)
$2K(x+1)-4K(y-3)+3K\ast z-2)=0$
$K\left[2\right(x+1)-4(y-3)+3(z-2\left)\right]=0$
$2x+2-4y+12+3z-6=0$
$2x-4y+3z+8=0$
This is the required equation.