Concept:Use the identity (a−b)2+(b−c)2+(c−a)2=2(a2+b2+c2−ab−bc−ca) and then (a+b+c)2=a2+b2+c2+2(ab+bc+ca).Explanation:Start with the given: (a−b)2+(b−c)2+(c−a)2=6.Apply the identity: this equals 2(a2+b2+c2−ab−bc−ca).So 6=2(a2+b2+c2−ab−bc−ca) → 3=a2+b2+c2−(ab+bc+ca).We know a2+b2+c2=29, thus 3=29−(ab+bc+ca) → ab+bc+ca=26.Now use (a+b+c)2=a2+b2+c2+2(ab+bc+ca).Substitute: (a+b+c)2=29+2×26=29+52=81.Take square root: a+b+c=±9.Answer:±9, which corresponds to option A.