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

Jul 28, 2018

ONE

#### Explanation:

$3 {x}^{2} + 3 {y}^{2} + {z}^{2} - 2 x y + 2 y z = 0$

$\implies 2 {x}^{2} + {y}^{2} + \left({x}^{2} - 2 x y + {y}^{2}\right) + \left({y}^{2} + 2 y z + {z}^{2}\right) = 0$

$\implies 2 {x}^{2} + {y}^{2} + {\left(x - y\right)}^{2} + {\left(y + z\right)}^{2} = 0$

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

$x = 0 , y = 0 , x = y \mathmr{and} y = - z$

Hence $x = y = z = 0$

So only ordered triplet is $\left(0 , 0 , 0\right)$

So the number of ordered triplet is ONE