What's the value of $a^2+b^2+c^2$ ?

Adding all three equations, we get $a^2+6a+b^2+2b+c^2+4c=-14$ .

Completing the squares, $a^2+6a+9+b^2+2b+1+c^2+4c+4=(a+3)^2+(b+1)^2+(c+2)^2=0$ .

If $a,b,c$ are real, this is only possible when $a=-3$ , $b=-1$ , $c=-2$ , so that $a^2+b^2+c^2=\boxed{14}$ .

However, there are other solutions if $a,b,c$ are allowed to be complex, which give different (complex) values for $a^2+b^2+c^2$ . One such solution is $a = 2 - i \sqrt3$ , $b = 3 + 2 i \sqrt3$ , $c = -1 - 3i \sqrt3$ , giving $a^2+b^2+c^2 = -28 + 14 i \sqrt3$ .

$a^2 + 2b = 7$

$b^2 + 4c = -7$

$c^2 + 6a = -14$

$\implies$

$a^2 + 2b + 1 = 8$

$b^2 + 4c + 4 = -3$

$c^2 + 6a + 9 = -5$

$\implies$

$(a + 3)^2 + (b + 1)^2 + (c + 2)^2 = 0 \implies a = -3, b = -1$ and $c = -2$ which are the only real values which satisfy the initial equations

$\therefore a^2 + b^2 + c^2 = \boxed{14}$ .