We have, 

$a x+y+z=a$   $...\left(\mathrm{i}\right)$

$x+by+z=b$   $...\left(\mathrm{ii}\right)$

$x+y+cz=c$   $...\left(\mathrm{iii}\right)$

are inconsistent. 

$\begin{array}{ll}\therefore & \left|\begin{array}{lll}a& 1& 1\\ 1& b& 1\\ 1& 1& c\end{array}\right|=0 \mathrm{and} \left|\begin{array}{lll}a& 1& 1\\ b& b& 1\\ c& 1& c\end{array}\right|\ne 0\end{array}$

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

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

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

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

and $\left|\begin{array}{lll}a& 1& 1\\ b& b& 1\\ c& 1& c\end{array}\right|\ne 0$

It is possible only when $a\ne 1,b\ne 1$ and $c\ne 1$