CGPET 2011 Solved Paper

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−‌

7

4

)x+(−3+‌

7

4

)y+(1−‌

7

4

)z
‌‌+(−4−‌

7

4

)=0
⇒x−5y−3z−23=0