Firstly we get that the equations are:
$$ab+c=2$$
$$ac+b=2$$
$$bc+a=2$$
Now notice that
$$(ba+c)-(bc+a)=2-2 \Rightarrow b(a-c)-(a-c)=0 \Rightarrow (b-1)(a-c)=0$$
This gives us 2 cases namely $b=1$ or $a=c$
\\Case 1: $b=1$
We can substitute this into the original equations that we had
$$a+c=2 \Rightarrow c=2-a$$
$$ac=1$$
Hence
$$a(2-a)=1 \Rightarrow 0=a^2 -2a +1 = (a-1)^2$$
This gives us the solutions $a=1$,$c=1$
Hence 1 set of solutions $(a,b,c)$ is
$$(1,1,1)$$
\\Case 2: $a=c$
We can substitute this into the original equations that we had
$$ab+a=2 \Rightarrow b=\frac{2}{a}-1$$
$$a^2+b=2$$
Hence
$$a^2 + \frac{2}{a}-1 = 2 \Rightarrow a^{3} - 3a +2 = 0$$
Since $a=1$ is a root of the equation $(a-1)$ is a factor
$$a^3 -3a +2 = (a-1)(a^2+a-2) = (a-1)(a-1)(a+2) = 0$$
Using $a=1$ gives us the same set of solutions as before but $a=-2$ gives us a new set, namely
$$(-2,-2,-2)$$
\\Further extension:
$$ab+c=k$$
$$ac+b=k$$
$$bc+a=k$$
Now notice that
$$(ba+c)-(bc+a)=k-k \Rightarrow b(a-c)-(a-c)=0 \Rightarrow (b-1)(a-c)=0$$ so we get the same equation in the general case
\\Case 1: $b=1$
We can substitute this into the original equations that we had
$$a+c=k \Rightarrow c=k-a$$
$$ac=k-1$$
Hence
$$a(k-a)=k-1 \Rightarrow 0=a^2 -ka +k-1 = (a-1)(a-(k-1))$$
This gives us the solutions $a=1$, $c=k-1$ (and $a=k-1$,$c=1$
Hence 1 set of solutions $(a,b,c)$ is
$$(1,1,(k-1))$$
but due to the symmetry of the problem all 3 permutations of this set are solutions, namely $((k-1),1,1)$, $(1,(k-1),1)$ as well.
\\Case 2: $a=c$
We can substitute this into the original equations that we had
$$ab+a=k \Rightarrow b=\frac{k}{a}-1$$
$$a^2+b=k$$
Hence
$$a^2 + \frac{k}{a}-1 = k \Rightarrow a^3 - (k+1)a +k = 0$$
Since $a=1$ is a root of the equation $(a-1)$ is a factor
$$a^3 -(k+1)a +k = (a-1)(a^2+a-k) = (a-1)(a-\phi)(a+(\phi+1)) = 0$$
$$\phi (\phi +1) =k \Rightarrow \phi^2+\phi-k$$
$$\phi = \frac{-1+\sqrt{4k+1}}{2}$$
$$-(\phi +1) = \frac{-1-\sqrt{4k+1}}{2}$$
\\Now we have new 2 new sets of solutions
When $a=\frac{-1+\sqrt{4k+1}}{2}$ gives us the set of solutions $$\left( \frac{-1+\sqrt{4k+1}}{2},\frac{-1+\sqrt{4k+1}}{2},\frac{-1+\sqrt{4k+1}}{2} \right)$$
When $a=\frac{-1 -\sqrt{4k+1}}{2}$ gives us the set of solutions
$$\left( \frac{-1-\sqrt{4k+1}}{2},\frac{-1-\sqrt{4k+1}}{2},\frac{-1-\sqrt{4k+1}}{2} \right)$$
Remark: To find $b$ we could substitute the value of $a$ into
$$b=\frac{k}{a} - 1$$
For the case where $a=\frac{-1+\sqrt{4k+1}}{2}$:
$$b= \frac{2k}{-1+\sqrt{4k+1}} -1 =\frac{2k+1-\sqrt{4k+1}}{-1+\sqrt{4k+1}}$$
$$=\frac{2k+1-\sqrt{4k+1}}{-1+\sqrt{4k+1}} \times \frac{1+\sqrt{4k+1}}{1+\sqrt{4k+1}} = \frac{2k+1-\sqrt{4k+1}+(2k+1)\sqrt{4k+1} -4k-1}{4k+1-1}$$
$$=\frac{-2k-\sqrt{4k+1}+(2k+1)\sqrt{4k+1}}{4k} = \frac{-2k+2k\sqrt{4k+1}}{4k} = \frac{-1+\sqrt{4k+1}}{2} = a$$
Which is why in fact $a=b=c$ (A very similar rationalizing of the denominator works for the case where $a=\frac{-1-\sqrt{4k+1}}{2}$)
Extension 1:
After having done the further extension for the general case we can substitute in $k=6$ to get the sets $(5,1,1)$ (and its permutations), $(2,2,2)$, $(-3,-3,-3)$
Extension 2:
$$ab-c=2$$
$$ac-b=2$$
$$bc-a=2$$
Now notice that
$$(ba-c)-(bc-a)=2-2 \Rightarrow b(a-c)+(a-c)=0 \Rightarrow (b+1)(a-c)=0$$
This gives us 2 cases namely $b=-1$ or $a=c$
\\Case 1: $b=-1$
We can substitute this into the original equations that we had
$$-a-c=2 \Rightarrow c=-2-a$$
$$ac=1$$
Hence
$$a(-2-a)=1 \Rightarrow 0=a^2 +2a +1 = (a+1)^2$$
This gives us the solutions $a=-1$,$c=-1$
Hence 1 set of solutions $(a,b,c)$ is
$$(-1,-1,-1)$$
Case 2: $a=c$
We can substitute this into the original equations that we had
$$ab-a=2 \Rightarrow b=\frac{2}{a}+1$$
$$a^2-b=2$$
Hence
$$a^2 - \frac{2}{a}-1 = 2 \Rightarrow a^{3} - 3a -2 = 0$$
Since $a=-1$ is a root of the equation $(a+1)$ is a factor
$$a^3 -3a -2 = (a+1)(a^2-a-2) = (a+1)(a+1)(a+2) = 0$$
Using $a=-1$ gives us the same set of solutions as before but $a=2$ gives us a new set, namely
$$(2,2,2)$$