Correct Answer - 1
Since vectors are coplanar .
`:. |{:(a,,1,,1),(1,,b,,1),(1,,1,,c):}|=0`
Applying `R_(2) to R_(2) -R_(1) ,R_(3) to R_(3) -R_(1),`
`|{:(a,,1,,1),(1-a,,b-1,,0),(1-a,,0,,c-1):}|=0`
`|{:(a//(1-a),,1//(1-b),,1//(1-c)),(1,,-1,,0),(1,,0,,-1):}|=0`
`rArr (a)/(1-a) (1)/(1-b) (-1) +(1)/(1-c)(1)=0`
`rArr (a)/(1-a) +(1)/(1-b) +(1)/(1-c) =0`
` rArr -1 +(1)/(1-a) +(1)/(1-b)+(1)/(1-c) =0`
`rArr (1)/(1-a) +(1)/(1-b) +(1)/(1-c)=1`