If $2^x =4^y=8z$ and $\bigg( \large\frac{1}{2x}+\frac{1}{4y}+\frac{1}{6z}\bigg)=\frac{24}{7}$ then the value of z is

Question

$\begin{array}{1 1} \large\frac{7}{16} \\ \large\frac{7}{32} \\ \large\frac{7}{48} \\ \large\frac{7}{64} \end{array} $