Explanation

This exploits the sum-of-squares identity (a + b + c)² − 3(ab + bc + ca) = (1/2)[(a − b)² + (b − c)² + (c − a)²], which is always non-negative. The given equation then reads (1/2)[(a − b)² + (b − c)² + (c − a)²] + 3(x² + y² + z²) = 0, a sum of non-negative terms equal to zero. Every squared term must vanish, so a − b = b − c = c − a = 0 and x = y = z = 0. Hence option (c): a = b = c and x = y = z = 0. Distractor (a) wrongly allows non-zero x, y, z, and (b) forces a = b = c = 0 although any common value works.