Let a , b , c be three distinct real numbers, none equal to one. If the vectors a i ^ + j ^ + k ^ , i ^ + b j ^ + k ^ and i ^ + j ^ + c k ^ are coplanar, then 1 1 - a + 1 1 - b + 1 1 - c is equal to

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TestHub · Accepted Answer

Given that the vectors are coplanar.

That means the determinant is equal to zero.

$\Rightarrow |\begin{matrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{matrix}| = 0$

$\Rightarrow R_{1} \to R_{1} - R_{2}$ and $R_{2} \to R_{2} - R_{3}$

$\Rightarrow |\begin{matrix} a - 1 & 1 - b & 0 \\ 0 & b - 1 & 1 - c \\ 1 & 1 & c \end{matrix}| = 0$

Expand the determinant along $C_{1}$ .

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

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

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

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

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

Hence this is the required option.

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