If $(x+y-7)[z(x+y)+24]=(y+z-7)[x(y+z)+24]=(z+x-7)[y(z+x)+24]$, find $x^2+y^2+z^2$ 
Let x, y, z be pairwise distinct real numbers, if $$(x+y-7)[z(x+y)+24]=(y+z-7)[x(y+z)+24] $$ $$=(z+x-7)[y(z+x)+24]$$, find $x^2+y^2+z^2$

I've tried many ways but couldn't find a working way to solve it. I tried letting $(x+y-7)[z(x+y)+24] = k$ and $t = x + y + z$ but none of these gives useful transformation as far as I can see. Could somebody shed some lights on this? Thanks in advance.

let $f_z=(x+y-7)(z(x+y)+24)$,$f_y,f_x$ defined similarly $$f_x=f_y=f_z=k$$ $$f_z-f_y=0\to -(y - z) (x^2 - 7 x - y z - 24)=0$$  so $$x^2-7x-yz=24$$and similarly for others yield  $$p_x=p_y=p_z=24$$ where $p_x=x^2-7x-yz$ and others defined similarly $$p_x-p_y=0\to (x-y)(x+y+z-7)=0$$ so $$x+y+z=7\to z=7-x-y$$ Plug this into orginal condition to get $$z^3-7z^2-24z-k=x^3-7x^2-24x-k=y^3-7y^2-24y-k=0$$ thus $x,y,z$ are the roots of cubic $$p(t)=t^3-7t^2-24t-k$$ Now by vieta you can easily find $x^2+y^2+z^2$
