Try $x = \frac{1}{2}, \; y = \frac{1}{3}, \; z = \frac{1}{6}.$ Your product is $ (2-1)(3-1)(6-1) = 1 \cdot 2 \cdot 5 = 10. $
Next, try Try $x = \frac{4}{7}, \; y = \frac{2}{7}, \; z = \frac{1}{7}.$ Your product is $ (\frac{7}{4}-1)(\frac{7}{2}-1)(7-1) = \frac{3}{4} \cdot \frac{5}{2} \cdot 6 = \frac{45}{4}. $
Your question has no fixed answer.
Take $x = \frac{10}{15}, \; y = \frac{3}{15}, \; z = \frac{2}{15}.$ Your product is $13.$
Take $x = \frac{6}{9}, \; y = \frac{2}{9}, \; z = \frac{1}{9}.$ Your product is $14.$
Take $x = \frac{15}{20}, \; y = \frac{3}{20}, \; z = \frac{2}{20}.$ Your product is $17.$
Take $x = \frac{14}{21}, \; y = \frac{6}{21}, \; z = \frac{1}{21}.$ Your product is $25.$
Take $x = \frac{165}{252}, \; y = \frac{77}{252}, \; z = \frac{10}{252}.$ Your product is $29.$
Take $x = \frac{21}{28}, \; y = \frac{6}{28}, \; z = \frac{1}{28}.$ Your product is $33.$
Take $x = \frac{65}{78}, \; y = \frac{10}{78}, \; z = \frac{3}{78}.$ Your product is $34.$
Take $x = \frac{35}{50}, \; y = \frac{14}{50}, \; z = \frac{1}{50}.$ Your product is $54.$
Take $x = \frac{85}{102}, \; y = \frac{15}{102}, \; z = \frac{2}{102}.$ Your product is $58.$
Take $x = \frac{170}{294}, \; y = \frac{119}{294}, \; z = \frac{5}{294}.$ Your product is $62.$
Take $x = \frac{270}{297}, \; y = \frac{22}{297}, \; z = \frac{5}{297}.$ Your product is $73.$
Take $x = \frac{77}{99}, \; y = \frac{21}{99}, \; z = \frac{1}{99}.$ Your product is $104.$
Take $x = \frac{247}{364}, \; y = \frac{114}{364}, \; z = \frac{3}{364}.$ Your product is $125.$
Take $x = \frac{90}{126}, \; y = \frac{35}{126}, \; z = \frac{1}{126}.$ Your product is $130.$
It seems likely that the target 16 requires at least two of $x,y,z$ to be irrational. Certainly you can fix, for example, $x = 1/2$ and solve for $y,z.$ So, take $x = \frac{1}{2}, \; \; y = \frac{1}{4} + \frac{1}{4} \sqrt{\frac{7}{15}}, \; \; z = \frac{1}{4} - \frac{1}{4} \sqrt{\frac{7}{15}}.$ Your product is $16.$