4. If the real numbers $x, y, z$ satisfy the system of equations:
$$
\left\{\begin{array}{l}
\frac{x y}{x+2 y}=1, \\
\frac{y z}{y+2 z}=2, \\
\frac{z x}{z+2 x}=3,
\end{array}\right.
$$

then ( ).
(A) $x+2 y+3 z=0$
(B) $7 x+5 y+2 z=0$
(C) $9 x+6 y+3 z=0$
(D) $10 x+7 y+z=0$
4. D.

From equations (1) and (3), we get $y=\frac{x}{x-2}, z=\frac{6 x}{x-3}$, hence $x \neq 0$.
Substituting into equation (2), we solve to get $x=\frac{27}{10}$.
Therefore, $y=\frac{27}{7}, z=-54$.
Verification shows that this set of solutions satisfies the original system of equations.
Thus, $10 x+7 y+z=0$.