nx  +  y  =  1  
ny  +  z  =  1  
x  +  nz  =  1 
If n is a real number, then the simultaneous system to the right has no solution if and only if n is equal to:
 1
 0
 1
 0 or 1
 ½
nx  +  y  =  1  
ny  +  z  =  1  
x  +  nz  =  1  


nx + y + ny + z + x + nz  =  3 
(n + 1)(x + y + z) = 3
This can be true unless n = 1.
The answer is (a).