If xy + yz + zx = 0, then ( 1x^2 - yz + 1y^2 - zx + 1z^2 - xy ) ( x,y,z 0 ) = ?

Answer: Option D

$$\begin{array}{rl}& xy+yz+zx=0\\ & \therefore xy+zx=-yz\\ & \Rightarrow xy+yz=-zx\\ & \Rightarrow yz+zx=-xy\\ & \therefore \frac{1}{x^2-yz}+\frac{1}{y^2-zx}+\frac{1}{z^2-xy}\end{array}$$
Putting values of -yz, -zx, -xy from above
$$\Rightarrow \frac{1}{x^2+\left(xy+zx\right)}+\frac{1}{y^2+\left(xy+yz\right)}$$       $$+\frac{1}{z^2+\left(yz+zx\right)}$$
$$\Rightarrow \frac{1}{x\left(x+y+z\right)}+\frac{1}{y\left(x+y+z\right)}$$       $$+\frac{1}{z\left(x+y+z\right)}$$
$$\begin{array}{rl}& \Rightarrow \frac{1}{\left(x+y+z\right)}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\\ & \Rightarrow \frac{1}{\left(x+y+z\right)}\left(\frac{zy+xz+xy}{xyz}\right)\\ & \Rightarrow \frac{1}{x+y+z}\times 0\\ & \Rightarrow 0\end{array}$$