Question

The equation of the plane passing through the point (1,1,1) and perpendicular to the planes 2x+y-2z=5 and 3x-6y-2z=7 is

• A. 14x + 2y - 15z = 1
• B. -14x + 2y +15z =3
• C. 14x - 2y + 15z = 27
• D. 14x + 2y + 15z = 31

Answer 1

The correct option is D

14x + 2y + 15z = 31

Let the equation of plane be ax + by + cz = 1. Then
$a+b+c=12a+b-2c=03a-6b-2c=0\Rightarrow a=7b,c=\frac{15b}{2}b=\frac{2}{31},a=\frac{14}{31},c=\frac{15}{31}\therefore 14x+2y+15z=31$