Find the value of $m$ if $\frac{m-a^{2}}{b^{2}+c^{2}}+\frac{m-b^{2}}{a^{2}+c^{2}}+\frac{m-c^{2}}{b^{2}+a^{2}}=3$ 
If $\dfrac{m-a^{2}}{b^{2}+c^{2}}+\dfrac{m-b^{2}}{a^{2}+c^{2}}+\dfrac{m-c^{2}}{b^{2}+a^{2}}=3;\ \ m,a,b,c \in\mathbb{R}$
Then the value of $m$ is...
Options 
$\boldsymbol{1.)}\ a^{2}-b^{2}-c^{2} \quad  \quad
\boldsymbol{2.)}\ a^{2}+b^{2}-c^{2}\\
\boldsymbol{3.)}\ a^{2}+b^{2} \quad  \quad   \quad   \quad
\boldsymbol{4.)}\ a^{2}+b^{2}+c^{2}$

By observation if all the three value 
$\dfrac{m-a^{2}}{b^{2}+c^{2}}=\dfrac{m-b^{2}}{a^{2}+c^{2}}=\dfrac{m-b^{2}}{a^{2}+c^{2}}=1$
Then $m=a^{2}+b^{2}+c^{2}$
I want to know if their is any other simple and short method other than the stated observation .
I have studied maths up to $12$th grade.

Plugging in $a = b = c$, we get
\begin{align*}
3 = \frac{m-a^{2}}{b^{2}+c^{2}}+\frac{m-b^{2}}{a^{2}+c^{2}}+\frac{m-c^{2}}{b^{2}+a^{2}}= 3\frac{m - a^2}{2a^2}
&\implies m - a^2 = 2a^2 \\
&\implies m = 3a^2.
\end{align*}
On the other hand, plugging $a = b = c$ into the answers we get
\begin{align*}
&\boldsymbol{(1)} \quad  - a^2 \\
&\boldsymbol{(2)} \quad a^2 \\
&\boldsymbol{(3)} \quad 2 a^2 \\
&\boldsymbol{(4)} \quad 3 a^2 \\
\end{align*}
so the answer must be $\boxed{\boldsymbol{(4)}}$.
