17. Find the real roots of the following equations:
$$
\begin{array}{l}
x^{2}+v^{2}+w^{2}=a^{2}, v w+u(y+z)=b c, \\
y^{2}+w^{2}+u^{2}=b^{2}, w u+v(z+x)=c a, \\
z^{2}+u^{2}+v^{2}=c^{2}, u v+w(x+y)=a b .
\end{array}
$$
$$
\text { Solve } \begin{array}{c}
\left(y^{2}+w^{2}+u^{2}\right)\left(z^{2}+u^{2}+v^{2}\right) \\
=b^{2} c^{2}=(v w+u y+u z)^{2} 。
\end{array}
$$

Simplifying, we get
$$
\left(u^{2}-y z\right)^{2}+(w u-v y)^{2}+(u v-w z)^{2}=0 \text { . }
$$

Since the roots are real, it is necessary that
$$
u^{2}-y z=0, w u-v y=0, u v-w z=0 .
$$

From other equations, similar results can be obtained.
Thus, $u^{2}=y z, v^{2}=2 x, w^{2}=x y$, $v w=u x, w u=v y, u v=w z$. Substituting these values into the known equations, we get
$$
\begin{array}{l}
x(x+y+2)=a^{2}, \quad u(x+y+z)=b c \\
y(x+y+z)=b^{2}, \quad v(x+y+z)=c a \\
z(x+y+z)=c^{2}, \quad w(x+y+z)=a b
\end{array}
$$

Therefore, $(x+y+z)^{2}=a^{2}+b^{2}+c^{2}$;
Hence, $x= \pm \frac{a^{2}}{\sqrt{a^{2}+b^{2}+c^{2}}}$,
$$
u= \pm \frac{b c}{\sqrt{a^{2}+b^{2}+c^{2}}} .
$$
$y$ and $z$ can be obtained similarly.