Pressure is isotropic
$$ F_{upthrust}=m_{dis}g $$
Intuitive proof: fluid surrounding the object does not “know” whether the fluid has been removed or is still there. Thus, remaining fluid will exert the same force as before on whatever object displaced the fluid.
Surface energy (at boundary between fluid and air)
$$ U=\gamma A $$
where $\gamma$ is the force per unit length of surface tension (or, from above, work per unit area associated with the boundary)
For a film of liquid, the force required to pull apart a rectangle is equal to $2\gamma l$ as there are two liquid-air interfaces.
For a solid-liquid-air interface, we have 3 terms representing the 3 types of boundaries.
$$ U = \gamma_{sl} A_{sl} + \gamma_{sv} A_{sv} + \gamma_{lv} A_{lv} $$
And the contact angle satisfies
$$ \cos\theta=\frac{\gamma_{sv}-\gamma_{sl}}{\gamma_{lv}} $$
which is trivial from the principle of virtual work. But we can also analyse this in terms of forces on the liquid wedge close to the boundary.
There are two $\gamma_{lv}$ terms as that is, by definition, the actual intermolecular forces between the liquid molecules. There is a rightwards force created by the solid of magnitude
$$ A_{sl}=\gamma_{sv}+\gamma_{lv}-\gamma_{sl} $$
As the solid-liquid interface is lower in energy than the two surfaces separated. Finally, there is a downwards force to keep the liquid from flying away.