The correct option is
C 2x+8y−8z−1=0Plane1 :3x+4y−5z+1=0Plane2 :5x+12y−12z=0
let us construct a ||gm ABCD with AB & AD in direction of normal to plane ⊥ & plane2 respectively.
−−→AB=3^i+4^j−5^k−−→AD=5^i+12^j−13^k
∴−−→AC will be the acute angle bisector whereas −−→BD will be in direction of obtuse angle bisector to the normals.
−−→AC=−−→AB+−−→AD (by ||gm law of addition )
−−→BD=−−→AB−−−→AD (by △ law of addition)
∴−−→BD=−2^i−8^j+8^k is the direction of the normal to the plane through obtuse angle bisector plane1 & plane2.
∴ Equation of plane through the line of intersection of plane1 & plane2
(3x+4y−5z+1)+λ(5x+12y−13z)=0(3+5λ)x+(4+12λ)y+(−5−13λ)+1=0
The above plane should be parallel to the plane formed as it is normal.
∴3+5λ−2=4+12λ−8=−5−13λ8⇒λ=−1
∴ The required plane is ,
−2x−8y+8z+1=0⇒2x+8y−8z−1=0