If 111abca3b3c3=(a−b)(b−c)(c−a)(a+b+c), where a, b, c are all different, then the determinant 111(x−a)2(x−b)2(x−c)2(x−b)(x−c)(x−c)(x−a)(x−a)(x−b) vanishes when

We have111abca3b3c3=(a−b)(b−c)(c−a)(a+b+c)           (1)Also, 111abca3b3c3=abc1a1b1c111a2b2c2 (taking a, b, c common from R1, R2, R3)=bcacab111a2b2c2  (Multiplying R1 by abc)=111a2b2c2bcacabThen   D=111(x−a)2(x−b)2(x−c)2(x−b)(x−c)(x−c)(x−a)(x−a)(x−b)=(a−b)(b−c)(c−a)(3x−a−b−c)Now, given that a, b and c are all different. So, D = 0 whenx=13(a+b+c)