From the given plane equations, π1+λλ2=0 (x+3y−6)+λ(3x−y+4z)=0 (1+3λ)x+(3−λ)y+4λz=0 ....(I) The perpendicular distance from (0,0,0) to above plane is 1 so,
|ax1+by1+cz1+d|
√a2+b2+c2
=1
|−6|2
√(1+3λ)2+(3−λ)2+(4λ)2
=1 1+9λ2+6λ+9+λ2−6λ+16λ2=16 26λ2=26 λ=±1 Then equation (I) gives, 4x+2y+4z−6=0 2x+y+2z−3=0