Equation of plane through the intersection of planes x+2y+3z−4=0 and 2x+y−z+5=0 is (x+2y+3z−4)+k(2x+y−z+5)=0 or (1+2k)x+(2+k)y+(3−k)z+(5k−4)=0.......(i) D.R′s of normal of plane (i) are =<(1+2k),(2+k),(3−k)> Given, 5x+3y+6z+8=0 ............(ii) D.R′s of plane (ii) are <5,3,6>. Since Eq.(i) is Perpendicular to the plane (ii), ∴5(1+2k)+(2+k)3+6(3−k)=0 ⇒5+10k+6+3k+18−6k=0 ⇒7k+29=0⇒k=
−29
7
∴ Required equation of plane is (x−2y+3z−4)+(−