Torques are always calculated about some specified reference point, if you change the reference point, you change the individual torque which each force correlates to (``produces''?). If the sum of the torques about a certain point is non-zero, then changing the point may change the total torque, but if the sum of torques about one point is zero (and the sum of the forces is zero), the sum of torques about any point will be zero. Notice, that it's the sum of torques which is zero.
Now to address your first diagram, if you calculate the torques about point A, then you must consider the interaction of the walls of the hole with the rod. Those interactions produce the torques which make the sum equal to zero. If the rod is in static equilibrium, you can't ignore the effects of the walls of the hole. And you can't say that the entire hole is a point.
Edit: If the rod is secured to the wall by attaching it to the outside by sending a screw or nail through the rod, then the friction between the wall and rod, or the friction between the rod and screw (or nail) will exert a torque about A.
Regarding the second diagram, if support C is touching the rod, it exerts a force on the rod and therefore has a torque about point B. If a different point (away from B) is chosen and the rod is touching support B, then B exerts a torque. But if the system is in rotational equilibrium, the sum of the torques about a chosen point must be zero.
In mechanics no. Torque is not a fundamental quantity. it's only job is to describe where in space a force is acting through (the line of action). Torque just describes a force at a distance. Given a force $\boldsymbol{F}$ and a torque $\boldsymbol{\tau}$ you can tell that the force acts along a line in space with direction defined by $\boldsymbol{F}$, but location defined by $\boldsymbol{\tau}$ as follows $$ \boldsymbol{r} = \frac{ \boldsymbol{F} \times \boldsymbol{\tau} }{ \| \boldsymbol{F} \|^2 } $$
In fact, you can slide the force vector anywhere along its line and it won't change the problem, so the $\boldsymbol{r}$ calculated above happens to be the point on the line closest to the origin.
It might be easier to discuss angular momentum first, since torque is the time derivative of angular momentum, just as force is the time derivative of linear momentum.
For a single particle with linear momentum $\boldsymbol{p} = m\boldsymbol{v}$ located at some instant at a point $\boldsymbol{r}$ the angular momentum is $$ \boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}$$
So where is the momentum line in space? The momentum line is called the axis of percussion. It is located at
$$ \boldsymbol{r} = \frac{ \boldsymbol{p} \times \boldsymbol{L} }{ \| \boldsymbol{p} \|^2 } = \frac{\boldsymbol{p} \times ( \boldsymbol{r} \times \boldsymbol{p})}{\| \boldsymbol{p} \|^2} = \frac{ \boldsymbol{r} (\boldsymbol{p} \cdot \boldsymbol{p}) - \boldsymbol{p} ( \boldsymbol{p} \cdot \boldsymbol{r}) }{\| \boldsymbol{p} \|^2} = \boldsymbol{r} \frac{ \| \boldsymbol{p} \|^2}{\| \boldsymbol{p} \|^2} = \boldsymbol{r} \; \checkmark $$
provided that the point $\boldsymbol{r}$ is perpendicular to the momentum $\boldsymbol{p}$. Let me elaborate. Imagine the direction of the line being $\boldsymbol{\hat{e}} = \boldsymbol{p} / \| \boldsymbol{p} \|$, and consider a point $\boldsymbol{r} + t \boldsymbol{\hat{e}}$ for some arbitrary scalar $t$. The angular momentum will be $\boldsymbol{L} = ( \boldsymbol{r} + t \boldsymbol{\hat{e}}) \times \boldsymbol{p} = \boldsymbol{r} \times \boldsymbol{p} $. So where along the line (the value of $t$) doesn't matter. Finally, if $\boldsymbol{r}$ is not perpendicular to $\boldsymbol{p}$ you can always find a value of $t$ that makes the point perpendicular. Set $t = -(\boldsymbol{r} \cdot \boldsymbol{p}) / \| \boldsymbol{p} \|$ and the point will be perpendicular.
Such a point can always be found, and it is the point on the line closest to the origin.
The conservation law for angular momentum (coupled with the conservation law for linear momentum) just states that not only the magnitude and direction of momentum is conserved but also the line in space where moment acts through is also conserved. So not only which direction is momentum point, by where is space it exists.
To visualize this, consider a case where you want to remove the momentum of a freely rotating body that is moving in space. You have a hammer, and you need to find out the following in order to completely stop the body. a) how much momentum to hit it with (the magnitude), b) in which direction to swing (direction) and c) where to hit it (location).
In summary, the common quantities in mechanics are interpreted as follows
$$ \begin{array}{r|l|l}
\text{concept} & \text{value} & \text{moment}\\
\hline \text{rotation axis} & \text{rot. velocity}, \boldsymbol{\omega} & \text{velocity}, \boldsymbol{v} = \boldsymbol{r}\times \boldsymbol{\omega} \\
\text{line of action} & \text{force}, \boldsymbol{F} & \text{torque}, \boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F} \\
\text{axis of percussion} & \text{momentum}, \boldsymbol{p} & \text{ang. momentum}, \boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}
\end{array} $$
The stuff under the value column are fundamental quantities that give us the magnitude of something (as well as the direction). The stuff under the moment column are secondary quantities that depend on where they are measured and give use the relative location of the fundamental quantities. Hence the terms torque = moment of force, velocity = moment of rotation and angular momentum = moment of momentum. All that means is that these quantities are $\boldsymbol{r} \times \text{(something fundamental)}$ and they describe the moment arm to this something.
The location of the line in space is always the same formula
$$ \text{(location)} = \frac{ \text{(value)} \times \text{(moment)}}{ \text{(magnitude)}^2} $$
where $\text{(magnitude)}$ is always the magnitude of the $\text{(value)}$ vector.
In statics, for example, we learn to balance forces and moments, which should be interpreted as balancing the force magnitude and the line of action of the force.
Best Answer
Varignon's theorem is applicable only for concurrent forces. The quoted part is incorrect and has been fixed (). For non-concurrent forces, the point of application of resultant force will change according to the moment.
In fact, the principle based on which we find the new resultant force's point of application is based on the fact that resultant moment should be the same as earlier. For couples, Moment has to be considered separately as their resultant could have no point of application which will yield the same moment (this is because resultant becomes 0 for a couple).