Question

find cartesian eqn & vector eqn of the planes passing thru the intersection of the planes r(2i+6j)+12=0 and r(3i-j+4k)=0 which is at unit distance from origin

Answer 1

your answer is here

Equation of given planes are 

r.(2i+6j)+12=0 ---(1) and r.(3i-j+4k)=0---(2)

Now, Let the Equation of plane passing through the intersection of these planes be 

r.(2i+6j)+12+λ[r.(3i-j+4k)]=0

=> r.{(2+3λ)i+(6-λ)j+4λk}+12=0 ---(3)

Distance of this plane from the origin , 

12/√(2+3λ)2+(6-λ)2+(4λ)2=1

Squaring the Given Expression

144= 4+9λ2+12λ+36+λ2-12λ+16λ2

=> 144-40= 26λ2

=> λ2=104/26

=> λ2=4

=> λ=+- 2

Putting the values in Equation (3), we get 

r.{(8)i+(4)j+8k}+12=0  and r.{(-4)i+(8)j-8k}+12=0