jmc

Since $x,y,z\ge 0$, we have by the AM-GM inequality that $$\sqrt[3]{xyz}\le \frac{x+y+z}{3}=\frac{7}{3}.$$Thus $xyz\le \frac{7^3}{3^3}=\frac{343}{27},$ with equality at $x=y=z=\frac{7}{3}.$

Also, $xyz\ge 0,$ so the range is $\boxed{\left[0,\frac{343}{27}\right]}$.