smc

Let $y=xr,z=x{r}^2$. Since $x,2y,3z$ are an arithmetic sequence, there is a common difference and we have $2xr-x=3x{r}^2-2xr$. Dividing through by $x$, we get $2r-1=3r^2-2r$ or, rearranging, $(r-1)(3r-1)=0$. Since we are given $x\ne y⟹r\ne 1$, the answer is $\boxed{\frac{1}{3}}$