For any Ax = 0, the non-trivial solution of x implies that |A| = 0
i.e $\begin{vmatrix} a &1 &1 \\ 1 &b &1 \\ 1&1 &c \end{vmatrix}$ = 0
Now looking into the qsn. we need some kind of (1-a), (1-b) & (1-c)
So, applying c1 = c1 - c2 & c3= c2 - c3 on the determinant we get,
$\begin{vmatrix} -(1-a) &1 &0 \\ 1-b &b &-(1-b) \\ 0&1 &1-c \end{vmatrix}$
so, -b(1-a)(1-c) - (1-a)(1-b) - (1-b)(1-c) = 0
or, b(1-a)(1-c) + (1-a)(1-b) + (1-b)(1-c) = 0
or, $\frac{b}{1-b} + \frac{1}{1-c} + \frac{1}{1-a}$ = 0 [ Divide both side by (1-a)(1-b)(1-c) ]
or, $\frac{1+b-1}{1-b} + \frac{1+c-c}{1-c} + \frac{1+a-a}{1-a}$ = 0
or, $\frac{b-1}{1-b} + \frac{1}{1-b} + \frac{1}{1-c} + \frac{1}{1-a}$ = 0
or, $\frac{1}{1-b} + \frac{1}{1-c} + \frac{1}{1-a}$ = 1