Let a ,   b ,   c be three distinct real numbers, none equal to one. If the vectors a i ^ + j ^ +…

Given that the vectors are coplanar.

That means the determinant is equal to zero.

$\Rightarrow \left|\begin{array}{ccc}a& 1& 1\\ 1& b& 1\\ 1& 1& c\end{array}\right|=0$

$\Rightarrow R_1\to R_1–R_2$ and $R_2\to R_2–R_3$

$\Rightarrow \left|\begin{array}{ccc}a-1& 1-b& 0\\ 0& b-1& 1-c\\ 1& 1& c\end{array}\right|=0$

Expand the determinant along $C_1$.

$\Rightarrow (a-1) \left[c\right(b-1)-(1-c\left)\right]+1\left[\right(1-b) (1-c\left)\right]=0$

$\Rightarrow c(a-1) (b-1)-(a-1) (1-c)+(1-b) (1-c)=0$

$\Rightarrow c(1-a) (1-b)+(1-a) (1-c)+(1-b) (1-c)+\left(1-a\right)\left(1-b\right)\left(1-c\right)=\left(1-a\right)\left(1-b\right)\left(1-c\right)$

$\Rightarrow (1-a) (1-b)+(1-a) (1-c)+(1-b) (1-c)=\left(1-a\right)\left(1-b\right)\left(1-c\right)$

$\Rightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$

Hence this is the required option.