4. If $x, y, z$ are real numbers, satisfying
$$
x+\frac{1}{y}=2 y+\frac{2}{z}=3 z+\frac{3}{x}=k \text{, and } x y z=3 \text{, }
$$

then $k=$ . $\qquad$
4. 4 .

Multiplying the three expressions yields
$$
\begin{array}{l}
k^{3}=6\left[x y z+\left(x+\frac{1}{y}\right)+\left(y+\frac{1}{z}\right)+\left(z+\frac{1}{x}\right)+\frac{1}{x y z}\right] \\
=6\left(3+k+\frac{k}{2}+\frac{k}{3}+\frac{1}{3}\right) \\
\Rightarrow k^{3}-11 k-20=0 \\
\Rightarrow(k-4)\left(k^{2}+4 k+5\right)=0 \\
\Rightarrow k=4
\end{array}
$$