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.

$(a-b)^2+(b-c)^2+(c-a)^2=2(3+4+5)=24$
Now $a-b=\dfrac1{a+b+c}$ etc.
Putting the values of $a-b, b-c, c-a;$
$$\frac1{(a+b+c)^2}(1^2+2^2+1^2)=24$$
$$\implies(a+b+c)^2=\frac14$$
$$\implies a+b+c=\pm\frac12$$
