For a non-trivial we must have
$\begin{vmatrix} 1& a& a\\b&1& b\\c& c& 1\end{vmatrix}=0$
Apply $C_1-C_2,C_2-C_3$ we get
$\begin{vmatrix} 1-a& 0& a\\b-1&1-b& b\\0& c-1& 1\end{vmatrix}=0$
$(1-a)[1-b-c(c-1)]+a(b-1)(c-1)=0$
$\large\frac{1}{c-1}+\frac{b}{b-1}+\frac{a}{a-1}=$$0$
$(\large\frac{1}{c-1}$$+1)+\frac{b}{b-1}+\frac{a}{a-1}$$=1$
$\large\frac{c}{c-1}+\frac{b}{b-1}+\frac{a}{a-1}$$=1$
$\large\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}$$=-1$
Hence (A) is the correct answer.