If x+y+z=0 then what is the value of $$\frac{x^{2}}{3z}+\frac{y^3}{3xz}+\frac{z^2}{3x}?$$

Given : $$x+y+z=0$$ ----------(i)

We know that : $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2+xy+yz+xz)$$

Using equation (i), => $$x^3+y^3+z^3-3xyz=0$$

=> $$x^3+y^3+z^3=3xyz$$ -----------(ii)

$$\therefore$$ $$\frac{x^{2}}{3z}+\frac{y^3}{3xz}+\frac{z^2}{3x}$$

= $$\frac{x^{3}}{3xz}+\frac{y^3}{3xz}+\frac{z^3}{3xz}=\frac{x^3+y^3+z^3}{3xz}$$

Multiplying both numerator and denominator by $$'y'$$

= $$\frac{y(x^3+y^3+z^3)}{3xyz}$$

Using equation (ii), => $$\frac{y(3xyz)}{3xyz}=y$$

=> Ans - (C)