<s>[INST] What can be the possible value of $a+b+c$ in the following case? What can be the possible value of $a+b+c$ in the following case?
$$a^{2}-bc=3$$ 
$$b^{2}-ca=4$$
$$c^{2}-ab=5$$
$0, 1, -1$ or $1/2$?

After doing $II-I$, $III-I$ and $III-II$, I got, 
$$(a+b+c)(b-a)=1$$
$$(a+b+c)(c-a)=2$$
$$(a+b+c)(c-b)=1$$
I'm unable to solve further, please help.
 [/INST] You have
$$\left\{
\begin{align*}
&(a+b+c)(b-a)=1\\
&(a+b+c)(c-a)=2\\
&(a+b+c)(c-b)=1\;.
\end{align*}\right.\tag{1}$$
These clearly imply that $a+b+c\ne 0$, so the first and third of these imply that $b-a=c-b$. In other words, $\langle a,b,c\rangle$ is an arithmetic progression. (The second equation of $(1)$ confirms this.) Set $d=b-a$; then $b=a+d$ and $c=a+2d$, so $a+b+c=3(a+d)$, and each of the equation in $(1)$ reduces to $3d(a+d)=1$.
Going back to the original equations, we see that
$$\begin{align*}
3&=a^2-bc=a^2-(a+d)(a+2d)\\
&=-3ad-2d^2=-3d(a+d)+d^2\\
&=d^2-1\;,
\end{align*}$$
or $d^2=4$. Thus, $d=\pm 2$, $1=3d(a+d)=\pm6(a+d)$, and $a+b+c=3(a+d)=\pm\frac12$.
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