If `|1 1 1a b c a^3b^2c^3|=(a-b)(b-c)(c-a)(a+b+c),w h e r ea ,b ,c` are different, then the determinant `|1 1 1(x-a)^2(x-b)^2(x-c)^2(x-b)(x-c)(x-c)(x-a)(x-a)(x-b)|` vanishes when `a+b+c=0` b. `x=1/3(a+b+c)` c. `x=1/2(a+b+c)` d. `x=a+b+c`