Solve for $x,y,z$ from the linear equations. The main question is :
$$\begin{align}
(b+c)(y+z)-ax &= b-c  \tag{1} \\
(c+a)(z+x)-by &= c-a  \tag{2} \\
(a+b)(x+y)-cz &= a-b  \tag{3}\\
\end{align}$$
Solve for $x,y,z$ if $a+b+c\ne0$
My method :
Opening all brackets, we get,
$$by+bz+cy+cz-ax=b-c$$
$$cz+cx+az+ax-by=c-a$$
$$ax+ay+bx+by-cz=a-b$$
Adding,
$$(a+b+c)(x+y+z)=0$$
Thus,
$$(x+y+z)=0$$
Also, 
$$by+bz+cy+cz-b+c=by$$
Thus,
$$b(y+z-1)+c(y+z+1)=ax$$
Similarly,
$$c(z+x-1)+a(z+x+1)=by$$
$$a(x+y-1)+b(x+y+1)=cz$$
I can't go any further, nothing is clicking. Please help me.

You're almost there; nice work!
Plug $x+y+z = 0$, in the form $(y+z) = -x$, into your first equation to get $(-x)(a+b+c) = b-c,$ so $$x = \frac{c-b}{a+b+c}.$$ 
Do similar things to the other two equations. 
