Suppose a,b,c ∈Rand abc,α≠0 If the system of equationsa+αx+αy+αz=0-----(1)αx+b+αy+αz=0-----(2)αx+αy+α+cz=0------(3) has a non-trivial solution, then  α1a+1b+1cis=

From eqns 1 &2 we haveax−by=0⇒ax=byFrom eqns 2 &3 we haveax−cz=0⇒ax=czax=by=cz⇒x/1/a=y/1/b=z/1/cSubstituting x,y,z values in equation (1), we get a+αa+αb+αc=0=1 ⇒ α1a+1b+1c=−1