Given : \( \Large 2^{x} = 4^{y} = 8^{z} \)

=> \( \Large 2^{x} = 2^{2y} = 2^{3z} \)

=> \( \Large x = 2y = 3z = k\ (say) \)

From \( \Large \frac{1}{2x}+\frac{1}{4y}+\frac{1}{6z}=\frac{24}{7} \)

\( \Large \frac{1}{2k}+\frac{1}{2k}+\frac{1}{2k}=\frac{24}{7} \)

=> \( \Large \frac{3}{2k}=\frac{24}{7} \)

=> \( \Large k = \frac{7}{16} \)

\( \Large \therefore z = \frac{k}{3} = \frac{7}{16 \times 3} = \frac{7}{48} \)