It is given, y(x+z)=19 y cannot be 19. If y=19,x+z=1 which is not possible when both x and z are natural numbers. Therefore, y=1 and x+z=19 It is given, z(x+y)=51 z can take values 3 and 17 Case 1: If z=3,y=1 and x=16 xyz=3×1×16=48 Case 2: If z=17,y=1 and x=2 xyz=17×1×2=34 Minimum value xyz can take is 34