Given a = x/(y - z), b = y/(z - x), and c = z/(x - y), where x, y, and z are not all zero, then the value of ab + bc + ca is

a=x/(y−z)⇒x−ay+az=0          (1)b=y/(z−x)⇒bx+y−bz=0          (2)c=z/(x−y)⇒−cx+cy+z=0        (3)Since x, y, z are not all zero, the above system has a non-trivial solution. So,Δ=1−aab1−b−cc1=0or    1 + ab + bc + ca = 0