Determine the real numbers $x, y, z, t$ satisfying an equation Be non-zero real numbers $a, b, c$ so $a + b + c \neq 0$ and $ab + bc + ca = 0$.
Determine the real numbers $x, y, z, t$ knowing that
$$\frac{\frac{a^2 + b^2 + c^2}{2}+x^2+y^2+z^2}{a+b+c}= t+ \sqrt{ax+by+cz-t(a+b+c)}.$$
I tried converting equality data but managed to find the numbers.
I wonder: is even possible to determine the numbers $x, y, z, t$ ?


is even possible to determine the numbers x, y, z, t?

Yes, it's possible.
Let $x^2+y^2+z^2=P,a+b+c=Q$.
Since
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=a^2+b^2+c^2$$
we have
$$\frac{\frac{a^2 + b^2 + c^2}{2}+x^2+y^2+z^2}{a+b+c}= t+ \sqrt{ax+by+cz-t(a+b+c)}$$
$$\Rightarrow\quad \frac{\frac{Q^2}{2}+P}{Q}= t+ \sqrt{ax+by+cz-Qt}$$
$$\Rightarrow\quad Q^2+2P-2Qt=2Q\sqrt{ax+by+cz-Qt}$$
Squaring the both sides gives
$$\Rightarrow\quad (Q^2+2P-2Qt)^2=4Q^2(ax+by+cz-Qt)$$
$$\Rightarrow\quad Q^4+4P^2+4Q^2t^2+4PQ^2-4Q^3t-8PQt=4Q^2(ax+by+cz)-4Q^3t$$
$$\Rightarrow\quad Q^4+4P^2+4Q^2t^2+4PQ^2-8PQt=4Q^2(ax+by+cz)$$
$$\Rightarrow\quad 4Q^2t^2-8PQt+Q^4+4P^2+4PQ^2-4Q^2(ax+by+cz)=0$$
$$\Rightarrow\quad t=\frac{2P\pm Q\sqrt{-(a-2x)^2-(b-2y)^2-(c-2z)^2}}{2Q}$$
Hence, in order for $t\in\mathbb R$ to exist, we have to have
$$a-2x=b-2y=c-2z=0,$$
i.e.
$$\color{red}{x=\frac a2,\quad y=\frac b2,\quad z=\frac c2}$$
and so $$t=\frac{P}{Q}=\frac{(a/2)^2+(b/2)^2+(c/2)^2}{a+b+c}=\frac{(a+b+c)^2}{4(a+b+c)}=\color{red}{\frac{a+b+c}{4}=t}$$
