The equation of a plane passing through the intersection of the planes 2x−3y+z−4=0 and x−y+z+1=0 is (2x−3y+z−4)+λ(x−y+z+1)=0 ⇒(2+λ)x+(−3−λ)y+(1+λ)z\+λ−4=0 . . . (i) But plane (i) is perpendicular to the plane ‌x+2y−3z+6=0 ‌∴(2+λ)1+(−3−λ)2+(1+λ)(−3)=0 ‌⇒‌‌2+λ−6−2λ−3−3λ=0 ‌⇒‌‌−4λ−7=0 ‌⇒‌‌λ=‌
−7
4
‌ Putting this value of λ in Eq. (i), we get the required equation of plane (2−‌