For given homogenous equation to have non-zero solution $ \begin{bmatrix} a & 1 & 1 \\ 1 & b & 1 \\1 & 1 & c \end{bmatrix}=0$
$R_1 \to R_2 -R_1$
$R_3 \to R_3 -R_1$
$ \begin{bmatrix} a & 1 & 1 \\ 1-a & b-1 & 0 \\1-a & 0 & c-1 \end{bmatrix}=0$
=> $a(b-1)(c-1) -1((1-a)(c-1))+1[0-(b-1)(1-a)]=0$
=> $a(b-1)(c-1) -1((1-a)(c-1))-(b-1)(1-a)=0$
=> $a(b-1)(c-1)+(a-1)(c-1))+(b-1)(a-1)=0$
Divide by $(1-a)(1-b)(1-c)$
$\large\frac{a(b-1)(c-1)}{(1-a)(1-b)(1-c)}+\frac{(a-1)(c-1)}{(1-a)(1-b)(1-c)}+\frac{(b-1)(a-1)}{(1-a)(1-b)(1-c)}$$=0$
$\large\frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$$=0$
$\large\frac{-(1-a-1)}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$$=0$
$-\large\frac{1-a}{1-a}+\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$$=0$
$\large\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$$=1$
Hence 1 is the correct answer.