If a + b + c = 0, then one of the solution of `|(a-x,c,b),(c,b-x,a),(b,a,c-x)| = 0` is

Options

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

Answer

A, B, C

Explanation

Applying `R_(1)toR_(1)R_(2)+R_(3)` and taking
`a+b+c-x` common from the first row the
`(a+b+c-x)|{:(1,0,0),(c,b-x,a),(b,a,c-x):}|=0`
Applying `C_(2)toC_(2)-C_(1)` and `C_(3)toC_(3)-C_(5)` then
`(a+b+c-x)|{:(1,0,0),(c,b-x-c,a-c),(b,a-b,c-x-b):}|=0`
`impliesx^(x^(2)-a^(2)-b^(2)-c^(2)+ab+bc+ac)=0`
`impliesx=0+-sqrt((3)/(2)(a^(2)+b^(2)+c^(2)))`
`becausea+b+c=0`