Concept:Let the common ratio be k and express x, y, z in terms of k.Explanation:Let b+c−ax​=c+a−by​=a+b−cz​=k. Then x=k(b+c−a), y=k(c+a−b), z=k(a+b−c). Substitute into (b−c)x+(c−a)y+(a−b)z: (b−c)x=k(b−c)(b+c−a), (c−a)y=k(c−a)(c+a−b), (a−b)z=k(a−b)(a+b−c). Sum = k[(b−c)(b+c−a)+(c−a)(c+a−b)+(a−b)(a+b−c)]. Expand each term: (b−c)(b+c−a)=(b2−c2)−a(b−c), (c−a)(c+a−b)=(c2−a2)−b(c−a), (a−b)(a+b−c)=(a2−b2)−c(a−b). Add the b2, c2, a2 parts: (b2−c2)+(c2−a2)+(a2−b2)=0. Add the remaining parts: −a(b−c)−b(c−a)−c(a−b)=−ab+ac−bc+ab−ac+bc=0. Total sum inside bracket is 0. Hence the entire expression equals k⋅0=0.Answer:0