7. Let $x, y, z > 0$, and $x + 2y + 3z = 6$, then the maximum value of $xyz$ is $\qquad$
7. $6=x+2y+3z \geqslant 3 \cdot \sqrt[3]{x \cdot 2y \cdot 3z}$, so $xyz \leqslant \frac{4}{3}$, hence the maximum value of $xyz$ is $\frac{4}{3}$.