if $\frac{{\log }_{2}x}{4}=\frac{{\log }_{2}y}{6}=\frac{{\log }_{2}z}{3p}\text{ and }{x}^3{y}^{2}z=1$ then p equal to

$\frac{{\log }_{2}x}{4}=\frac{{\log }_{2}y}{6}=\frac{{\log }_{2}z}{3p}=q,(q\ne 0)$
${\log }_{2}x=4q,{\log }_{2}y=6q,{\log }_{2}z=3pq$
we get $x=2^{4q},y=2^{6q},z=2^{3pq}$ respectively
from $x^3{y}^{2}z=1$ we get
$\begin{array}{r}12q+12q+3pq=0\\ \Rightarrow 8+p=0\Rightarrow p=-8\end{array}$