Double Equality

Given $\begin{cases} \displaystyle \sum_{cyc} \dfrac 1a = \sum_{cyc} \dfrac 1{ab} & ... (1) \\ \displaystyle \sum_{cyc} (a+b)^2 = \sum_{cyc} (a+b-c)^2 & ... (2) \end{cases}$

From equation 1:

$\begin{aligned} \frac 1a + \frac 1b + \frac 1c & = \frac 1{ab} +\frac 1{bc} + \frac 1{ca} \\ \frac {ab+bc+ca}{abc} & = \frac {a+b+c}{abc} \\ \implies ab+bc+ca & = a+b+c = \color{#3D99F6} k \end{aligned}$

From equation 2:

$\begin{aligned}\sum_{cyc} (a+b)^2 & = \sum_{cyc} (a+b-c)^2 \\ & = \sum_{cyc} \left((a+b)^2-2c(a+b)+c^2 \right) \\ \implies \sum_{cyc} c^2 & = 2 \sum_{cyc} c(a+b) \\ a^2+b^2+c^2 & = 4(ab+bc+ca) = 4\color{#3D99F6}k \end{aligned}$

Now we have:

$\begin{aligned} (a+b+c)^2 & = a^2+b^2+c^2 + 2(ab+bc+ca) \\ k^2 & = 4k + 2k \\ k(k-6) & = 0 \\ \implies k & = \begin{cases} 0 \\ 6 \end{cases} \\ \implies a^2+b^2+c^2 & = 4k = \begin{cases} 0 \\ 24 \end{cases} \end{aligned}$

Therefore, the sum of all possible values of $a^2+b^2+c^2$ is $\boxed{24}$ .

Note that

$\begin{aligned} (a + b + c)^2 & = a^2 + b^2 + c^2 + 2(ab + ac + bc) \\ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} & = \frac{ab + ac + bc}{abc} \\ \frac{1}{ab} + \frac{1}{ac} + \frac{1}{bc} & = \frac{a + b + c}{abc} \\ (a + b)^2 + (a + c)^2 + (b + c)^2 & = 2(a^2 + b^2 + c^2) + 2(ab + ac + bc) \\ (a + b - c)^2 + (b + c - a)^2 + (c + a - b)^2 & = 3(a^2 + b^2 + c^2) - 2(ab + ac + bc) \end{aligned}$

Since the second and third expressions are equal, $a + b + c = ab + ac + bc$ . Substituting into the first equation, we have that $(ab + ac + bc)^2 - 2(ab + ac + bc) = a^2 + b^2 + c^2$ . Since the fourth and fifth expressions are equal, we have $a^2 + b^2 + c^2 = 4(ab + ac + bc)$ . This is a system of two equations and two unknowns, giving solutions $ab + ac + bc = 0,6$ and $a^2 + b^2 + c^2 = 0,24$ . Hence, the answer is $0 + 24 = \boxed{24}$ .