Question

The number of ordered triples (x,y,z) satisfy `3x^2+3y^2+z^2-2xy+2yz=0` is?

Answer 1

ONE

Explanation:

`3x^2+3y^2+z^2-2xy+2yz=0`

`=>2x^2+y^2+(x^2-2xy+y^2)+(y^2+2yz+z^2)=0`

`=>2x^2+y^2+(x-y)^2+(y+z)^2=0`

Hence for real values of `x,y,z` to satisfy the above relation we have

`x=0,y=0,x=yand y=-z`

Hence `x=y=z=0`

So only ordered triplet is `(0,0,0)`

So the number of ordered triplet is ONE