Maximizing the sum of utilities or more generally weighted sum of utilities will give you efficient allocations as solutions.
To get efficient allocations, we can solve the following problem:
$\max_{((x_1,y_1),(x_2,y_2),(x_3,y_3))\in\mathbb{R}^2_+\times\mathbb{R}^2_+\times\mathbb{R}^2_+}\alpha_1\min(x_1,y_1)+\alpha_2\min(2x_2,y_2)+\alpha_3\min(3x_3,y_3)$
subject to $x_1+x_2+x_3=\omega_X, \ y_1+y_2+y_3=\omega_Y$
where $(\alpha_1,\alpha_2,\alpha_3)\in\mathbb{R}^3_+\setminus (0,0,0)$ and $\omega_X>0$ and $\omega_Y>0$ are given.
Solution to the above problem is necessarily Pareto efficient. As we vary $(\alpha_1,\alpha_2,\alpha_3)$ and consider the union of set of solutions generated in the process, we get the set of all Pareto efficient allocations.
For example: maximising the sum of utilities for the case where $\omega_X=\omega_Y=\omega>0$,
$\max_{((x_1,y_1),(x_2,y_2),(x_3,y_3))\in\mathbb{R}^2_+\times\mathbb{R}^2_+\times\mathbb{R}^2_+}\min(x_1,y_1)+\min(2x_2,y_2)+\min(3x_3,y_3)$
subject to $x_1+x_2+x_3=\omega, \ y_1+y_2+y_3=\omega$
will give the following set of feasible allocations as solutions to the above problem:
$\{((x_1,y_1),(x_2,y_2),(x_3,y_3))\in\mathcal{F}|y_1\leq x_1 \wedge y_2 \leq 2x_2 \wedge y_3 \leq 3x_3\}$
Here $\mathcal{F}$ is the set of feasible allocations i.e.
$\mathcal{F}=\{((x_1,y_1),(x_2,y_2),(x_3,y_3))\in\mathbb{R}^2_+\times\mathbb{R}^2_+\times\mathbb{R}^2_+|x_1 +x_2+x_3=y_1+y_2+y_3=\omega\}$
These solutions are Pareto efficient, but not necessarily always the only efficient allocations. There can be more.
