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If the lines $a x+y+1=0, x+b y+1=0$ and $x+y+c=0 \quad(a, b, c$ being distinct and different from 1) are concurrent, then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=$
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The correct answer is:
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If the given lines are concurrent, then
$\left|\begin{array}{lll}
a & 1 & 1 \\
1 & b & 1 \\
1 & 1 & c
\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}
a & 1-a & 1-a \\
1 & b-1 & 0 \\
1 & 0 & c-1
\end{array}\right|=0$
$\left\{\right.$ Apply $C_2 \rightarrow C_2-C_1$ and $\left.C_3 \rightarrow C_3-C_1\right\}$
$\begin{aligned}
& \Rightarrow a(b-1)(c-1)-(b-1)(1-a)-(c-1)(1-a)=0 \\
& \Rightarrow \frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0 \\
& \Rightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1
\end{aligned}$
$\left|\begin{array}{lll}
a & 1 & 1 \\
1 & b & 1 \\
1 & 1 & c
\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}
a & 1-a & 1-a \\
1 & b-1 & 0 \\
1 & 0 & c-1
\end{array}\right|=0$
$\left\{\right.$ Apply $C_2 \rightarrow C_2-C_1$ and $\left.C_3 \rightarrow C_3-C_1\right\}$
$\begin{aligned}
& \Rightarrow a(b-1)(c-1)-(b-1)(1-a)-(c-1)(1-a)=0 \\
& \Rightarrow \frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0 \\
& \Rightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1
\end{aligned}$
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