In cartesian coordinates, we can derive the vector $\vec v_3$ by vector subtraction $\vec v_2-\vec v_1$. We then get the distance between $P$ och $Q$ by taking the absolute value of $\vec v_3$ which then is: $$\lvert \vec v_3\rvert = \lvert \vec v_2-\vec v_1 \rvert= \lvert (x_2,y_2,z_2)-(x_1,y_1,z_1) \rvert = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$
But how can we do this completely in cylindrical coordinates without converting to cartesian coordinates? We only have, and can only use, $\rho_1$ and $\theta_1$ for $\vec v_1$ as well as $\rho_2$ and $\theta_2$ for $\vec v_2$ in cylindrical coordinates $(\rho,\theta,z)$, where $z=0$ in this example.
Your problem for $z=0$ reduces to the polar coordinate problem already answered here. $$|\vec v_3|^2=(\vec v_2-\vec v_1)\cdot(\vec v_2-\vec v_1)=|\vec v_1|^2+|\vec v_2|^2-2\vec v_2\cdot \vec v_1$$ Now using cylindrical coordinates, $|\vec v_i|^2=r_i^2+z_i^2$ and $\vec v_2\cdot \vec v_1=r_1r_2\cos(\theta_2-\theta_1)+z_1z_2$. Then the final answer will be $$|\vec v_3|^2=r_1^2+r_2^2-2r_1r_2\cos(\theta_2-\theta_1)+(z_2-z_1)^2$$