Equsion

For the equations not to have a solution, the coefficient matrix $\begin{pmatrix} a & b & c\\ b & c & a\\ c & a & b \end{pmatrix}$

must be singular - ie its determinant must be zero; that is $3abc=a^3+b^3+c^3$

This equation is symmetric in $a,b,c$ , so it makes sense to define $u=a+b+c$ , $v=bc+ca+ab$ , $w=abc$ (symmetric polynomials). Rewriting the above equation, we find $u(u^2-3v)=0$

Since in the problem, $a,b,c$ are all positive, we can't have $u=0$ ; therefore $u^2=3v$ .

Expanding this out again, we get $a^2+b^2+c^2=bc+ca+ab$

We can treat this as a quadratic in the variable $c$ ; solving, we find $\begin{aligned} c&=\frac{a+b \pm \sqrt{(a+b)^2-4(a^2+b^2-ab)}}{2} \\ &= \frac{a+b \pm \sqrt{-3a^2+6ab-3b^2}}{2} \\ &= \frac{a+b \pm \sqrt3 \sqrt{-(a-b)^2}}{2} \end{aligned}$

This only has a real solution when $a=b=c$ . Hence the only value the sum in the question can take is $\boxed{\frac32}$ .

---

For the bonus question (where $a,b,c$ are real, non-zero and distinct - not necessarily positive), $u^2=3v$ is impossible (since $a,b,c$ are real and distinct); but we can now have $u=0$ . In this case, each term of the sum is $-1$ so the sum itself is $-3$ (for example, take $a=1,b=2,c=-3$ ).

Just set a=b=c and you'll get it. (BETTER SOLUTION NEEDED)