If `|[a,b-c,c+b],[a+c,b,c-a],[a-b,a+b,c]|=0` , then the line `a x+b y+c=0` posses through the fixed point which is

Options

A. `(1,2)`
B. `(1,1)`
C. `(-2,1)`
D. `(1,0)`

Answer

B

Explanation

Applying `C_(1) to aC_(1)` and then `C_(1) to C_(1) +bC_(2) +cC_(3)` and taking
`(a^(2)+b^(2)+c^(2))` common from `C_(1) ` we get
` Delta =((a^(2)+b^(2)+c^(2)))/(a) |{:(1,,b-c,,c+b),(1,,b,,c-a),(1,,b+a,,c):}|`
`=((a^(2)+b^(2)+c^(2)))/(a) |{:(1,,b-c,,c+b),(0,,c,,-a-b),(0,,a+c,,-b):}|`
`(R_(2) to R_(2)-R_(1),R_(3) to R_(3)-R_(1))`
`=((a^(2)+b^(2)+c^(2)))/(a) (-bc+a^(2)+ab+ac+bc)`
(expanding along `C_(1)`)
`=(a^(2)+b^(2)+c^(2))(a+b+c)`
hence `Delta =0 rArr a+b+c=0`
Therefore line ax+by+c=0 passes through the fixed point (1,1)