Equation of plane passing through (2,−3,4) having DR′s(a,b,c) is given by a⋅(x−2)+b(y+3)+c(z−4)=0 ∴⇒ax+by+cz−2a+3b−4c=0...(i Since, Eq. (i) is perpendicular to both the planes 2x−3y+5z=2 and x+y+2z=3 So, 2a−3b+5c=0 . . . (ii) and a+b+2c=0 . . . (iii) On solving Eqs. (i) and (ii), we get
a
−6−5
=
b
5−4
=
c
2+3
=k( say ) (a,b,c)=(−11k,k,5k) From Eq. (i), we get −11x+y+5z−2×(−11)+3×1−4×5=0 −11x+y+5z=−5 ⇒x−