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If vectors $a \hat{i}+\hat{j}+\hat{k}, \hat{i}+b \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+\hat{k}$ $(a \neq b \neq c \neq 1)$ are coplanar, then find $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$
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Since vectors are coplanar
$\therefore\left|\begin{array}{lll}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ccc}a & 1 & 1 \\ 1-a & b-1 & 0 \\ 0 & 1-b & c-1\end{array}\right|=0 \quad$ [Using $R_{2}-R_{1}$
$\left.\mathrm{R}_{3}-\mathrm{R}_{2}\right]$
$\Rightarrow a(b-1)(c-1)-(1-a)\{(c-1)-(1-b)\}=0$
$\Rightarrow a(1-b)(1-c)+(1-a)(1-c)+(1-a)(1-b)$
$=0$
$\Rightarrow(a-1+1)(1-b)(1-c)+(1-a)(1-c)$
$+(1-a)(1-b)=0$
$\Rightarrow(1-b)(1-c)+(1-a)(1-c)+(1-a)(1-b)$
$=(1-a)(1-b)(1-c)$
$\Rightarrow \quad \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$
$\therefore\left|\begin{array}{lll}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ccc}a & 1 & 1 \\ 1-a & b-1 & 0 \\ 0 & 1-b & c-1\end{array}\right|=0 \quad$ [Using $R_{2}-R_{1}$
$\left.\mathrm{R}_{3}-\mathrm{R}_{2}\right]$
$\Rightarrow a(b-1)(c-1)-(1-a)\{(c-1)-(1-b)\}=0$
$\Rightarrow a(1-b)(1-c)+(1-a)(1-c)+(1-a)(1-b)$
$=0$
$\Rightarrow(a-1+1)(1-b)(1-c)+(1-a)(1-c)$
$+(1-a)(1-b)=0$
$\Rightarrow(1-b)(1-c)+(1-a)(1-c)+(1-a)(1-b)$
$=(1-a)(1-b)(1-c)$
$\Rightarrow \quad \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$
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