8. Solve the system

$$
\left\{\begin{array}{l}
5 x^{2}+3 y^{2}+3 x y+2 x z-y z-10 y+5=0 \\
49 x^{2}+65 y^{2}+49 z^{2}-14 x y-98 x z+14 y z-182 x-102 y+182 z+233=0
\end{array}\right.
$$
Answer: $(0,1,-2),\left(\frac{2}{7}, 1,-\frac{12}{7}\right)$.

Consider the second equation of the system as a quadratic equation in terms of the difference $(x-z)$:

$$
49(x-z)^{2}-14(x-z)(y+13)+65 y^{2}-102 y+233=0 .
$$

$\frac{D}{4}=49(y+13)^{2}-49\left(65 y^{2}-102 y+233\right)=-(56)^{2}(y-1)^{2} \leqslant 0$.

Therefore, the second equation has a solution only if $y=1$. Then from the second equation, we get that $x-z=2$. Substituting $z=x-2$ and $y=1$ into the first equation, we obtain:

$$
5 x^{2}+3 x+2 x(x-2)-(x-2)-2=0 \Leftrightarrow 7 x^{2}-2 x=0 \Rightarrow x=0(z=-2), x=\frac{2}{7}\left(z=-\frac{12}{7}\right)
$$