If `|(1,1,1),(a,b,c),(a^(3),b^(3),c^(3))|= (a -b) (b -c) (c -a) (a + b+c)` where a, b, c are all 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 whenA. `a+b+c=0`B. `x=(1)/(3) (a+b+c)`C. `x=(1)/(2) (a+b+c)`D. `x=a+b+c`

If `|(1,1,1),(a,b,c),(a^(3),b^(3),c^(3))|= (a -b) (b -c) (c -a) (a + b

1.	If `\|(1,1,1),(a,b,c),(a^(3),b^(3),c^(3))\|= (a -b) (b -c) (c -a) (a + b+c)` where a, b, c are all 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 whenA. `a+b+c=0`B. `x=(1)/(3) (a+b+c)`C. `x=(1)/(2) (a+b+c)`D. `x=a+b+c`
Answer» Correct Answer - B we have `\|{:(1,,1,,1),(a,,b,,c),(a^(3),,b^(3),,c^(3)):}\|=(a-b)(b-c)(c-) (a+b+c)` `" Also " \|{:(1,,1,,1),(a,,b,,c),(a^(3),,b^(3),,c^(3)):}\|` `=abc \|{:((1)/(a),,(1)/(b),,(1)/(c)),(1,,1,,1),(a^(2),,b^(2),,c^(2)):}\|" ""(taking a,b,c common from "R_(1),R_(2),R_(3)")"` `=\|{:(bc,,ac,,ab),(1,,1,,1),(a^(2),,b^(2),,c^(2)):}\|` (Multiplying `R_(1)` by abc) `=\|{:(1,,1,,1),(a^(2),,b^(2),,c^(2)),(bc,,ac,,ab):}\|` `" Then " D = \|{:(1,,1,,1),((x-a)^(2),,(x-b)^(2),,(x-c)^(2)),((x-b)(x-c),,(x-c)(x-a),,(x-a)(x-b)):}\|` Now given that a,b and C are all different So D=0 when `x=(1)/(3) (a+b+c)`