If the vectors `ahati+hatj+hatk, hati+bhatj+hatk, hati+hatj+chatk(a!=1, b!=1,c!=1)` are coplanat then the value of `1/(1-a)+1/(1-b)+1/(1-c)` is (A) 0 (B) 1 (C) -1 (D) 2

Answer

`1`

Explanation

Given that the vectors `vecu = a hati + hatj +hatk`,
`vecv =hati +b hatj +hatk and vecw =hati +hatj +chatk`, where a, b, c `ne` 1 are coplanar. Therefore,
`|{:(a,,1,,1),(1,,b,,1),(1,,1,,c):}|=0`
Operating `C_1 to C_1 - C_2, C_2 to C_2 - C_3`
`|{:(a-1,,0,,1),(1-b,b-1,,1),(0,,1-c,,c):}|=0`
Expanding
`c(a-1) (b-1)+(1-b)(1-c)-(1-c)(a-1)=0`
`" "(c)/(1-c) + (1)/(1-a)+ (1)/(1-b) =0`
`" "(c)/(1-c) + 1+ (1)/(1-a)+ (1)/(1-b) =1`
`" " (1)/(1-c) + (1)/(1-a) + (1)/(1-b)=1`