Let's approach this problem by assuming the legs of the polygon are elastic and they carry as much force as they are deflected (compressed), but the table itself is a rigid body.
Theory
This means that for each location $\overline{r}_i = (x_i, y_i)$ the deflection is a function of the overall vertical deflection $\delta$ of the table and the two tilt angles $\overline{\theta} = ( \theta_x, \theta_y )$ in the following way
$$ \delta_i = \delta + \overline{\theta} \times \overline{r}_i $$
This expression is akin to the velocity kinematics of a rigid body $\vec{v}_i = \vec{v} + \vec{\omega} \times \vec{r}_i$, but applied for a small time slice such that $\overline{\theta} = \vec{\omega} \Delta t$ and $\delta = \vec{v} \Delta t$.
The 2D cross-product operator $\times$ results only in the in-plane component of the resulting vector. So the above is $\delta_i = \delta + \theta_x y_i - \theta_y x_i$.
At each leg the vertical force $F_i$ is calculated from the table deflection as
$$F_i = k \,\delta_i = k \left( \delta - \overline{r}_i \times \overline{\theta} \right) $$
Now to sum up all the forces and moments at the origin
$$ F_{\rm net} = \sum_i F_i $$
$$ \overline{M}_{\rm net} = \sum_i \overline{r}_i \times F_i$$
And again the cross product is appropriately defined such that $\overline{r}_i \times F_i = ( y_i F_i, -x_i F_i )$
The net force/moment at the origin, must be equal to the applied force/moment (Newton's 3rd law). If the location of the applied force is $\overline{c} = (x_c,y_c)$ and the total load is $W$, then
$$ \overline{M}_{\rm net} = \overline{c} \times W $$
$$ F_{\rm net} = W$$
Now to express the above relationships in terms of linear-algebra.
The leg forces are
$$\begin{bmatrix}F_{1}\\
F_{2}\\
\vdots\\
F_{n}
\end{bmatrix}=k\,\begin{bmatrix}y_{1} & -x_{1} & 1\\
y_{2} & -x_{2} & 1\\
\vdots & \vdots\\
y_{n} & -x_{n} & 1
\end{bmatrix}\begin{bmatrix}\theta_{x}\\
\theta_{y}\\
\delta
\end{bmatrix} \tag{1}$$
and the balance of forces are
$$\begin{bmatrix}y_{1} & y_{2} & \cdots & y_{n}\\
-x_{1} & -x_{2} & \cdots & -x_{n}\\
1 & 1 & & 1
\end{bmatrix}\begin{bmatrix}F_{1}\\
F_{2}\\
\vdots\\
F_{n}
\end{bmatrix}=\begin{bmatrix}y_{c}W\\
-x_{c}W\\
W
\end{bmatrix}$$
Now recognize the (n×3) common coefficient matrix
$$ {\rm G}=\begin{bmatrix}y_{1} & -x_{1} & 1\\
y_{2} & -x_{2} & 1\\
\vdots & \vdots\\
y_{n} & -x_{n} & 1
\end{bmatrix} \tag{2} $$
and normalize each force to a fraction of the applied load $w_i = F_i/W$ and re-write the above equation(s) after inserting (1) into (2) as
$$\left({\rm G}^{\intercal}k\,{\rm G}\right)\begin{bmatrix}\theta_{x}\\
\theta_{y}\\
\delta
\end{bmatrix}=\begin{bmatrix}y_{c}\\
-x_{c}\\
1
\end{bmatrix}W \tag{3}$$
with direct solution
$$\begin{bmatrix}\theta_{x}\\
\theta_{y}\\
\delta
\end{bmatrix}=\tfrac{W}{k}\left({\rm G}^{\intercal}{\rm G}\right)^{-1}\begin{bmatrix}y_{c}\\
-x_{c}\\
1
\end{bmatrix} \tag{4}$$
and in terms of the normalized leg forces, the stiffness and applied load all cancel out to get
$$\begin{bmatrix}w_{1}\\
w_{2}\\
\vdots\\
w_{n}
\end{bmatrix}={\rm G}\left({\rm G}^{\intercal}{\rm G}\right)^{-1}\begin{bmatrix}y_{c}\\
-x_{c}\\
1
\end{bmatrix} \tag{5}$$
Solution
So the process has three steps
- Define the coefficient matrix $$\small {\rm G}=\begin{bmatrix}y_{1} & -x_{1} & 1\\
y_{2} & -x_{2} & 1\\
\vdots & \vdots\\
y_{n} & -x_{n} & 1
\end{bmatrix}$$
- Define the constant vector $$\small {\rm C}=\begin{bmatrix}y_{c}\\
-x_{c}\\
1
\end{bmatrix}$$
- Calculate the normalized weights for each leg (the baryweights as they are called) $$\small {\rm w}={\rm G}\left({\rm G}^{\intercal}{\rm G}\right)^{-1}{\rm C}$$
Stability
Now some of the baryweights might be negative. This is ok, as long as there are at least 3 legs with positive weight on them.
You can go through the above steps and for any leg you find is in tension (negative weight), you can remove it as a row from ${\rm G}$ and try again, until a) you have only 3 positive weights, or b) have one negative weight and 3 legs which means the table is unstable and it will tip over.