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

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Answer 1 · 2018-07-28T04:10:55

ONE

#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