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