Equation of the plane passing through the line of intersection of x+y+z=1 and 2x+3y+4z=5 is (2x+3y+4z−5)+λ(x+y+z−1)=0 ⇒(2+λ)x+(3+λ)y+(4+λ)z+(−5−λ)=0 .......(i) ∵ plane(i) is perpendicular to the plane x−y+z=0 ∴(2+λ)(1)+(3+λ)(−1)+(4+λ)(1)=0 2+λ−3−λ+4+λ=0⇒λ=−3 Hence, equation of required plane is −x+z−2=0 or x−z+2=0 ⇒r.(