find the equation of the plane through the intersection of the planes 2x+6y+12=0 and 3x-y+4z=0 which are at a unit - Maths - Three Dimensional Geometry

Question

find the equation of the plane through the intersection of the
planes 2x+6y+12=0 and 3x-y+4z=0 which are at a unit distance from
the origin​

Rahul Raj · Accepted Answer

Dear student

the equation of the plane through the intersection of the planes
2x+6y+12=0 and 3x-y+4z=0 is

2x+6y+12+λ(3x-y+4z)=0

x(2+3λ)+y(6-λ)+4λz+12=0

As â€‹the plane is 
at a unit distance from the originâ€‹(0,0,0), we
have

0(2+3λ)+0(6-λ)+4λ*0+12(2+3λ)2+(6-λ)2+4λ2=112(2+3λ)2+(6-λ)2+4λ2=1(2+3λ)2+(6-λ)2+4λ2=14426λ2+40=14426λ2=104λ=±2Plane is 8x+4y+8z+12=0-4x+8y-8z+12=0

Regards

find the equation of the plane through the intersection of the planes 2x+6y+12=0 and 3x-y+4z=0 which are at a unit distance from the origin​