Extra conditioning

$\begin{aligned} a+b+c&={a}^2+{b}^2\\&={\left(a+b\right)}^2-2ab\\{\left(a+b\right)}^2-(a+b)&=2ab+c\\(a+b)(a+b-1)&=3c\\(c-2)(c-1)&=3c\because a+b+c=2\\{c}^2-6c+2&=0\\\Rightarrow \sum { c } &=6\\p&=ab+bc+ca\\&=ab+c(a+b)\\&=c+c(2-c)\\&=3c-{c}^2\\&=3c-q \because q=abc, ab=c\\p+q&=3c\\\sum{p+q}&=3\sum{c}\\&\huge\color{#3D99F6}{=\boxed{18}}\end{aligned}$

By Vieta's formula, we have:

$\begin{cases} a+b+c = 2 & ...(1) \\ ab+bc+ca = p & ...(2) \\ abc = q &...(3) \end{cases}$

It is given that:

$\begin{cases} a+b+c = a^2 + b^2 & ...(4) \\ ab = c & ...(5) \end{cases}$

Therefore, we have:

$\begin{aligned} (4): \quad \color{#3D99F6}{a+b+c} & = a^2 + b^2 \quad \quad \small \color{#3D99F6}{(1): \ \ a+b+c = 2} \\ \implies a^2 + b^2 & = \color{#3D99F6}{2} \quad ...(4a) \end{aligned}$

$\begin{aligned} (5): \quad ab & = c \\ \implies \color{#3D99F6}{abc} & = c^2 \quad \quad \small \color{#3D99F6}{(3): \ \ abc = q} \\ c^2 & = \color{#3D99F6}{q} \quad ...(5a) \end{aligned}$

$\begin{aligned} (2): \quad \color{#3D99F6}{ab}+bc+ca & = p \quad \quad \small \color{#3D99F6}{(5): \ \ ab = c} \\ \color{#3D99F6}{c}+c(\color{#D61F06}{a+b}) & = p \quad \quad \small \color{#D61F06}{(1): \ \ a+b+c = 2} \\ c+c(\color{#D61F06}{2-c}) & = p \\ c + 2c - \color{#3D99F6}{c^2} & = p \quad \quad \small \color{#3D99F6}{(5a): \ \ c^2 = q} \\ 3c - \color{#3D99F6}{q} & = p \\ \implies p + q & = 3c \quad ...(2a) \end{aligned}$

Now,

$\begin{aligned} (\underbrace{a+b+c}_{(1)})^2 & = \underbrace{a^2+b^2}_{(4a)} + c^2 + 2(\underbrace{ab+bc+ca}_{(2)}) \\ 4 & = 2 + c^2 +2\color{#3D99F6}{p} \quad \quad \small \color{#3D99F6}{(2a): \ \ p+q = 3c \implies p = 3c - q} \\ 2 & = c^2 +2(\color{#3D99F6}{3c-q}) \quad \quad \small \color{#3D99F6}{(5a): \ \ c^2=q} \\ 2 & = c^2 +6c - 2c^2 \\ \implies c^2 - 6c +2 & = 0 \\ c & = \begin{cases} c_1 = 3 + \sqrt{7} \\ c_2 = 3 - \sqrt{7} \end{cases} \end{aligned}$

Therefore, the sum all possible $p+q = 3c \quad ...(2a)$ :

$\begin{aligned} (p_1+q_1) + (p_2+q_2) & = 3c_1 + 3c_2 \\ & = 3(3 + \sqrt{7}) + 3(3 - \sqrt{7}) \\ & = \boxed{18} \end{aligned}$

From Vieta's formulas we know that $a+b+c=2$ then $a+b=2-c$ , $abc=q$ and $ab+bc+ac=p$ , thus, from the given equations, we have that $q=-c^2$ and $p=c+bc+ac=c(1+a+b)=c(3-c)=3c-c^2$ , therefore, the sum of all possibles values of $p+q$ is $c^2+3c-c^2=3c$ , that is, three times the sum of all possible values for $c$ .

Now, as $a+b=2-c$ and $ab=c$ , then $a+b=2-ab$ , which is the same that $2(a+b)=4-2ab$ . Also, $a^2+b^2=2$ , thus, $(a+b)^2=2+2ab$ . Adding both equations, $(a+b)^2+2(a+b)=6$ , and solving for $a+b$ , we conclude that $(a+b)=-1\pm \sqrt { 7 }$ , hence, $c=2-(a+b)=2+1\pm \sqrt { 7 }=3\pm \sqrt { 7 }$ , so the sum of all possible values for $c$ is $6$ , then, the asked value is $3\cdot 6=18$ .

Let us find value of $p+q$ . $p+q = ab(1+a+b+ab)$ .

By $Viete's Theoram$

$a+b+c=2=a+b+ab$ .

Therefore sum of all values of p+q are

$3\sum ab$ .

Now

$a+b=2-ab$ ......… …1

And

$a^{2}+b^{2} =2$ .

Forming a quadratic in $ab$ by squaring 1, we calculate sum of roots as 6, ie

$\sum ab$ = 6.

Thus answer is 18.