**The correct answer is option c) 1**

if the given lines are concurrent , then \( \begin{array}{|ccc|cc|} a& 1 & 1 \\ 1 & b & 1\\ 1 &1 & c\\ \end{array}=0 \)

Applying { \(C_2\rightarrow C_2-C_1 and \space C_3\rightarrow C_3-C_1\)}

\( \begin{array}{|ccc|cc|} a& 1 -a& 1 -a\\ 1 & b-1 & 0\\ 1 &0 & c-1\\ \end{array}=0 \)

= a(b-1)(c-1) -(c-1)(1-a)-(b-1)(1-a)=0

=\({a\over 1-a}+{1\over 1-b}+{1\over 1-c}\)

(dividing by (1-a )(1-b)(1-c))

\(={1\over 1-a}+{1\over 1-b}+{1\over 1-c}=1\)