(d) 0
Given, x + y + z = 0 ⇒ x + y = – z
⇒ x2 + y2 + 2xy = z2 ⇒ x2 + y2 = z2 – 2xy
∴ \(\frac{1}{x^2+y^2-z^2} = \frac{1}{z^2-2xy-z^2}=\frac{1}{-2xy}=-\frac{1}{2xy}\)
Similarly, \(\frac{1}{y^2+z^2-x^2} = -\frac{1}{2xy}\) and \(\frac{1}{z^2+x^2-y^2}=-\frac{1}{2zx}\)
∴ \(\frac{1}{x^2+y^2-z^2}+\frac{1}{y^2+z^2-x^2}+\frac{1}{z^2+x^2-y^2}\)
= \(-\frac{1}{2xy}-\frac{1}{2yz}-\frac{1}{2zx}\) = \(-\frac{1}{2}\big[\frac{z+x+y}{xyz}\big]\) = 0. [∵ x + y + z = 0]