xy+yz=0 y(x+z)=0 ⇒y=0 and (x+z)=0 ⇒y=0 is a equation of xz-plane. ⇒x+z is also an equation of plane. d.r′.s of normal to the plane y=0 is (0,1,0) and d.r′.s of normal to the plane x+z=0 is (1,0,1)
Now, a1a2+b1b2+c1c2=0×1+1×0+0×1=0
Hence, both planes are perpendicular.