Three Dimensional Geometry Part 2

Given planes,
x−y−z−1=0 ...(i)
2x+y−3z+4=0 ...(ii)
Equation of plane passing through line of intersection of planes (i) and (ii) is given by
(x−y−z−1)+λ(2x+y−3z+4)=0 Distance of plane (iii) from origin =

2

21

(given)
⇒

|4λ−1|

√(2λ+1)2+(λ−1)2+(−3λ+1)2

=√

2

21

Squaring both sides

(4λ−1)2

(2λ+1)2+(λ−1)2+(3λ+1)2

=

2

21

⇒21(16λ2−8λ+1)=2(14λ2+8λ+3)
⇒308λ2−184λ+15=0
308λ2−154λ−30λ+15=0
(2λ−1)(154λ−15)=0
⇒λ=

1

2

or λ=

15

154

Putting λ=

1

2

in Eq. (iii), we have
4x−y−5z+2=0