Find out........................

$(a+b+c)^{2}=0$

$a^{2}+b^{2}+c^{2}+2(ab+bc+ac)=0$

$a^{2}+b^{2}+c^{2}=-2(b(a+c)+ac)$

$a^{2}+b^{2}+c^{2}=-2(-b^{2}+ac)$

$a^{2}+b^{2}+c^{2}=2(b^{2}-ac)$

put in the equation and answer is 2

It is given that $a+b+c=0\quad \Rightarrow b = - (a+c)$ .

$\Rightarrow \dfrac {b^2+c^2+a^2}{b^2-ca} = \dfrac {[-(a+c)]^2+c^2+a^2}{[-(a+c)]^2-ca} = \dfrac {c^2+2ca+a^2+c^2+a^2}{c^2+2ca+a^2-ca}$

$= \dfrac {2c^2+2ca+2a^2}{c^2+ca+a^2} = \dfrac {2(c^2+ca+a^2)}{c^2+ca+a^2} = \boxed{2}$

let us assume that a=3,b=1,c=-4 so that it satisfies a+b+c=0 if we substitute the same a,b,c values in (b^2+c^2+a^2)/(b^2-ca) then we will get 2 as a result in case if we consider another values fora,b,c also then we will get 2 as the final result we can take a=4,b=1,c=-5; a=6,b=1,c=-7 .....................so on