NeosIII

I can't vouch for how good they are, but I found these:
 
https://github.com/lovasoa/whitebophir
 
https://github.com/BuchholzTim/Whitebird
 
If you're in an organization that uses Office 365, the virtual whiteboard built into Teams is pretty decent.

I've used OpenBoard before, and it does work pretty good for just pure whiteboard software. Problem is I don't know of a way to use it to modify from a different machine. Something like remote desktop or running it on an iPad and screen sharing it to the desktop could work, but I don't know how good a solution that would be for you. 

This is an example of minor edits to your original code doing it correctly:
 
#include <iostream>
#include <iomanip>
int main ()
{
    
    for (int i = 1; i <= 9; i += 3)
    {
        for (int j = 1; j <= 9; j++)
        {
            for (int k = j; k <= j; k++) 
            {
                std::cout << i << " x" << std::setw(2) << j << " =" << std::setw(2) << i * j << "   ";
                std::cout << i+1 << " x " << j << " =  " << (i+1)*j << " ";
                std::cout << i+2 << " x " << j << " = " << (i+2)*j << " ";
                std::cout << "\n";
            }
        }
        
        std::cout << "\n";
        
    }
}

Hey so I looked back over your code and there is some minor logic errors with what you're attempting to do. If you want them to be printed out like in the photos where 1 2 and 3's multiplication is all store in the same "table" then you need to account for that in your programming.
So you first loop needs to increment i by 3. Then you need to also account for the changes to i in the print out where you increase i by 2 and 3 as well when couting out. I editted your example to show a possible way of doing this.
 so for instance you need to change what happens inside your final for loop a bit.
thinking something along the lines of adding this to your code:
 
std::cout << i+1 << " x " << j << " =  " << (i+1)*j << " ";
It's more than just this since you want 3 answers attached. But it should hopefully be straight forward on how to get the 3rd number if you input this into your code.
 
I put the changes into your code and tested it on my end getting the result you were asking for. 1 is grouped with 2 and 3, 4 with 5 and 6 etc.
 
Sorry that I missed understood what you were asking for initially I quickly read through it.

So just a very quick look at it. You could use an if else statement that only sends out the endl or “\n” for every third one printed out. So you could cout each time it shoots out and at 2/3 depending on if you’d start counting at 0 or 1 either reset the count, or sent it up to do it only when the count is divisible by 3. I’m hoping that’s what you were asking for.