If a, b, c, x, y, z are real numbers such that (a+b+c)^{2}-3(ab+bc+ca)+3(x^{2}+y^{2}+z^{2})=0, then which one of the following is <strong>CORRECT</strong>?

  1. A. a=b=c, x=y=z \neq 0
  2. B. a=b=c=0, x=y=z=1
  3. C. a=b=c, x=y=z=0
  4. D. a \neq b \neq c, x=y=z=0

Correct Answer: C. a=b=c, x=y=z=0

Explanation

Expanding the equation gives a^2+b^2+c^2-ab-bc-ca + 3x^2+3y^2+3z^2 = 0. Multiplying by 2 yields (a-b)^2+(b-c)^2+(c-a)^2 + 6x^2+6y^2+6z^2 = 0. Since the sum of squares is zero, each term must be zero, giving a=b=c and x=y=z=0.

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