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 when

Options

A. `a+b+c=0`
B. `x=(1)/(3) (a+b+c)`
C. `x=(1)/(2) (a+b+c)`
D. `x=a+b+c`

Answer

B

Explanation

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)`