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|(a-b-c,2a,2a),(2b,b-c-a,2b),(2c,2c,c-a-...

`|(a-b-c,2a,2a),(2b,b-c-a,2b),(2c,2c,c-a-b)|`

A

`2 (a+b+c)`

B

`-2 (a+b+c)`

C

`-(a+b+c)`

D

`abc`

Text Solution

Verified by Experts

The correct Answer is:
B
`Delta|{:(a-b-c, 2a, 2a),(2b, b-c-a, 2b),(2c, 2c, c-a-b):}|`
`R_(1)to R_(1)+ R_(2)+ R_(3)`
`Delta|{:(a+b+c, a+b+c, a+b+c),(2b, b-c-a, 2b), (2c, 2c, c-a-b):}|implies Delta (a+b+c)|{:(1,1,1),(2b, b-c-a, 2b),(2c, 2c, c-a-b):}|`
Now `C_(1) to C_(1) -C_(2)`
`therefore Delta = (a+b+c) |{:(0,1,1),(a+b+c, b-c-a, 2b),(0, 2c, c-a-b):}|implies Delta = (a+b+c) ^(2) |{:(0,1,1),(1, b-c-a, 2b),(0, 2c, c-a-b):}|`
Expanding along 1st column
`Delta = (a+b+c )^(3){-(c-a-b-2c)}`
`Delta = ( a+b+c) ^(3) = (a+b+c) (a+b+c)^(2)`
Either `x =0 or x =-2 (a+b+c)`
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