If x, y, z are different from zero and Δ=ab−yc−za−xbc−za−xb−yc=0, then the value of the expression ax+by+cz is
0
-1
1
2
Applying R2→R2−R1,R3→R3−R1, we get
Δ=ab−yc−z−xy0−x0z=0
Expanding along C3 we get
(c−z)-xy-x0+zab−y−xy=0
or (c - z) (xy) + z(ay + bx - xy) = 0
or cxy - xyz + ayz + bxz - xyz = 0
or ayz + bzx + cxy = 2xyz
or ax+by+cz=2