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3 is the right answer using the identity
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then we have (x+y+z)²=1²
                   ⇒ x²+y²+c²+2(xy+xz+zy)=1
substituting value of xy+yz+zx
⇒ x²+y²+z²-2=1
⇒ x²+y²+z²=²1+2=3
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