<s>[INST] 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.
 [/INST] I write $u = y + z$, $v = z + x$, $w = x + y$ and $s = (u + v + w) / 2 (= x + y + z)$.
The equation becomes $$(u - 7)(us - u^2 + 24) = (v - 7)(vs - v^2 + 24) = (w - 7)(ws - w^2 + 24).$$
Taking the difference of the first and the second term, we get $$(u - v)((u + v - 7)s - (u^2 + uv + v^2 - 7(u + v) - 24)) = 0.$$
Since $x, y, z$ are all different, we know that $u, v, w$ are all different. This leads to $$(u + v - 7)s = u^2 + uv + v^2 - 7(u + v) - 24 \tag{1}$$ and symmetrically $$(v + w - 7)s = v^2 + vw + w^2 - 7(v + w) - 24.$$ Taking the difference of the above two identities, we get $$(u - w)s = (u - w)(u + w + v - 7)$$ which leads to $$s = 2s - 7$$ and thus $s = 7$.
Now (1) becomes $$u^2 + uv + v^2 - 14(u + v) + 25 = 0. \tag{2}$$ Taking the symmetric sum of (2), we get $$2\sum u^2 + \sum uv - 56s + 75 = 0. \tag{3}$$
If we write $t = \sum x^2$, then we have $\sum xy = (s^2 - t)/2 = (49 - t)/2$. It follows that $$\sum u^2 = \sum (y^2 + 2yz + z^2) = 2\sum x^2 + 2\sum xy = 49 + t$$ and $$\sum uv = ((2s)^2 - \sum u^2) / 2 = (147 - t) / 2.$$
Thus the equation (3) reads $$2(49 + t) + (147 - t) / 2 = 317$$ and we deduce that $t = 97$.

Finally, it is possible to give examples of $x, y, z$ satisfying these equalities.
E.g. $x = 3, y = 2 + 2\sqrt{10}, z = 2 - 2\sqrt{10}$.
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