Given `a=x//(y-z), b=y//(z-x),a n dc=z//(x-y),w h e r ex , y , za n dz` are not all zero, then the value of `a b+b c+c a` is `0` b. `1` c.`-1""` d. none of these

Options

A. 0
B. 1
C. -1
D. none of these

Answer

C

Explanation

`a=x//(y-z ) rArr x-ay+ ax=0`
`b=y //(z-x) rArr bx+ y-bz=0`
`c=z//(x-y) rArr -cx +cy +z=0`
Since x,y,z are not all zero , the above system has non-trivial solution .So
`Delta =|{:(1,,-a,,a),(b,,1,,-b),(-c,,c,,1):}|=0`
`" or " 1+ab +bc +ca =0`