11. Solve the system of equations in the set of real numbers $\left\{\begin{array}{l}x^{2}+y^{2}+z^{2}=\frac{9}{4} \\ -8 x+6 y-24 z=39 .\end{array}\right.$
11. By the Cauchy-Schwarz inequality, we have $\left(x^{2}+y^{2}+z^{2}\right)\left[(-8)^{2}+6^{2}+(-24)^{2}\right] \geqslant(-8 x+6 y-24 z)^{2}$.

Since $\left(x^{2}+y^{2}+z^{2}\right)\left[(-8)^{2}+6^{2}+(-24)^{2}\right]=\frac{9}{4} \times(64+36+4 \times 144)=39^{2}$,
$$
\begin{array}{l} 
\text { and }(-8 x+6 y-24 z)^{2}= 39^{2}, \\
\left(x^{2}+y^{2}+z^{2}\right)\left[(-8)^{2}+6^{2}+(-24)^{2}\right]=(-8 x+6 y-24 z)^{2},
\end{array}
$$

that is, the equality holds in inequality (1).
Thus, by the condition for equality in the Cauchy-Schwarz inequality, we get $\frac{x}{-8}=\frac{y}{6}=\frac{z}{-24}$.
Combining this with $-8 x+6 y-24 z=39$, we obtain $x=-\frac{6}{13}, y=\frac{9}{26}, z=-\frac{18}{13}$.