3. Given $x-\frac{a y z}{x^{2}}=y-\frac{a z x}{y^{2}}=z-\frac{a x y}{z^{2}}=\lambda$, and the real numbers $x, y, z$ are distinct. Then $x+y+z=(\quad)$.
(A) $\frac{\lambda}{a}$
(B) $a \lambda$
(C) $\lambda-a$
(D) $\lambda+a$
3. D.

From the given, we have
$$
x^{3}-a y z=\lambda x^{2}, \quad y^{3}-a z x=\lambda y^{2} \text {. }
$$

Subtracting the two equations, we get
$$
x^{3}-y^{3}+a z(x-y)=\lambda\left(x^{2}-y^{2}\right) \text {. }
$$

Since $x-y \neq 0$, we have
$$
\lambda(x+y)=x^{2}+x y+y^{2}+a z \text {. }
$$

Similarly, $\lambda(y+z)=y^{2}+y z+z^{2}+a x$.
Subtracting (1) from (2), we get
$$
\left.\lambda(x-z)=x^{2}-i^{2}+\cdots(x-z)-n t r-y\right)
$$

Since $x-z \neq 0$, we have $\lambda=x+y+z-a$, so $x+y+z=\lambda+a$.