Let a, b, c be three distinct real numbers, none equal to one. If...

Given that the vectors are coplanar.

That means the determinant is equal to zero.

⇒ \left|\begin{cases}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{cases}\right| = 0

$⇒ {R}_{1} \to {R}_{1} – {R}_{2}$ and ${R}_{2} \to {R}_{2} – {R}_{3}$

⇒ \left|\begin{cases}a - 1 & 1 - b & 0 \\ 0 & b - 1 & 1 - c \\ 1 & 1 & c\end{cases}\right| = 0

Expand the determinant along ${C}_{1}$.

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

$⇒ c ( a - 1 ) ( b - 1 ) - ( a - 1 ) ( 1 - c ) + ( 1 - b ) ( 1 - c ) = 0$

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

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

$⇒ \dfrac{1}{1 - a} + \dfrac{1}{1 - b} + \dfrac{1}{1 - c} = 1$

Hence this is the required option.