I couldn't resist, so: First, you should note:
- The tension in the string will be the same along its entire length (assuming frictionless conditions)
- To use the hooks optimally, you should arrange them that they carry equal load
Taking this into consideration, the angle that the string bends at each hook should be the same (the wall hooks should be spaced equally on the arc of a circle between the picture hooks, not a catenary as noted).
The picture below illustrates:
$$ø = \frac{2θ}{3}$$
Or more generally:
$$ø = \frac{2θ}{n}$$
where n = the number of hooks
Below is the FBD at one of the frame hooks:
- $F_t$ is the tension in the string.
- $F_p$ is the vertical force on the
frame hook.
From this, the tension in the string can be calculated as:
$$F_t = \frac{F_p}{sin(θ)}$$
The FBD at one of the hooks is as follows:
With $F_h$ as the resultant force on the hook.
$F_h$ can be calculated as:
$$F_h = 2F_tsin(0.5ø)$$
This can be generalised as:
$$F_h = \frac{2F_p}{sin(θ)}*sin(\frac{θ}{n})$$
From this you can see that as $θ$ approaches $0$, $F_h$ approaches $2F_p/n$, but $F_t$ approaches infinity. Thus you should consider the tensile strength of the string and the forces that the frame and frame hooks can handle as well.
You can hang your picture with 3 hooks with $θ = 62.11$ degrees, giving you $F_t = 28.285N$ and $F_h = 19.999N$. This will be "safe" assuming that the gravitational constant is 10, wihch it isn't, but that's just nit picking.