Question

The equation of the plane through the point $(2,5,-3)$ perpendicular to the planes $x+2y+2z=1$ and $x-2y+3z=4$ is:

• A. $3x-4y+2z-20=0$
• B. $7x-y+5z=30$
• C. $x-2y+z=11$
• D. $10x-y-4z=27$

Answer 1

The correct option is D $10x-y-4z=27$

Let the DR of the normal of the plane be $<a,b,c>$

$\therefore$ The plane is perpendicular to the plane $x+2y+2z=1$

$\therefore$ Sum of product of the DRs of the normal be zero

$\therefore a+2b+2c=0$ ...... $\left(1\right)$

It is also perpendicular to the plane

$x-2y+3z=4$

$\therefore$ Sum of the product of the DRs of the normal be zero

$\therefore a-2b+3c=0$ ...... $\left(2\right)$

$a+2b+2c=0$

$a-2b+3c=0$

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

$a=10,b=-1,c=-4$

The plane passes through point $(2,5,-3)$

$\therefore 10(x-2)+(-1)(y-5)-4(z+3)=0$

$\Rightarrow 10x-20-y+5-4z-12=0$

$\Rightarrow 10x-y-4z-27=0$

$10x-y-4z=27$