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

Question

Let $a,b,c$ be three distinct real numbers, none equal to one. If the vectors $a\hat{i}+\hat{j}+\hat{k},\hat{i}+b\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+c\hat{k}$ are coplanar, then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is equal to :

Accepted Answer

$\begin{vmatrix} a \quad  1 \quad  1 \ 1 \quad  b \quad  1 \ 1 \quad  1 \quad  c \end{vmatrix}=0$

${C}_{2}	o {C}_{2}-{C}_{1},{C}_{3}	o {C}_{3}-{C}_{1}$

$\begin{gathered} |\begin{array}{ccc}a \quad  1-a \quad  1-a \ 1 \quad  b-1 \quad  0 \ 1 \quad  0 \quad  c-1\end{array}|=0 \ a(b-1)(c-1)-(1-a)(c-1)+(1-a)(1-b)=0 \ a(1-b)(1-c)+(1-a)(1-c)+(1-a)(1-b)=0 \end{gathered}$

$\begin{gathered} \frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0 \ \Rightarrow -1+\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0 \ \Rightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1 \end{gathered}$

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