If the system of equations $ax+y+z=0,x+by+z=0$  and $x+y+cz=0\left(a,b,c\ne 1\right)$  has a non-trivial solution, then

$ax+y+z=0,x+by+z=0$  and $x+y+cz=0\left(a,b,c\ne 1\right)$  has a non-trivial solution

$\left|\begin{array}{ccc}a& 1& 1\\ 1& b& 1\\ 1& 1& c\end{array}\right|=0$

$a\left(bc-1\right)-1\left(c-1\right)+1\left(1-b\right)=0$

$abc-a-c+1+1-b=0$

$abc-a-b-c+2=0\Rightarrow a+b+c=2+abc$

$\frac{1}{a-1}+\frac{1}{b-1}+\frac{1}{c-1}=-1$

$\Rightarrow$ $\left(b-1\right)\left(c-1\right)+\left(a-1\right)\left(c-1\right)+\left(a-1\right)\left(b-1\right)=-\left(a-1\right)\left(b-1\right)\left(c-1\right)$

$\Rightarrow$ $bc-b-c+1+ac-a-c+1+ab-a-b+1=\left(ab-a-b+1\right)\left(1-c\right)$

$\Rightarrow$ $ab+bc+ca-2\left(a+b+c\right)+3=\left(ab-a-b+1-abc+ac+bc-c\right)$

$\Rightarrow$ $ab+bc+ca-2\left(a+b+c\right)+3=ab+bc+ac-a-b-c+1-abc$

$\Rightarrow$ $a+b+c=2+abc$