$9 \cdot 56$ Find the minimum value of
$$
\begin{aligned}
A & =\sqrt{\left(1264-z_{1}-\cdots-z_{n}\right)^{2}+x_{n}^{2}+y_{n}^{2}}+ \\
& \sqrt{z_{n}^{2}+x_{n-1}^{2}+y_{n-1}^{2}}+\cdots+\sqrt{z_{2}^{2}+x_{1}^{2}+y_{1}^{2}}+ \\
& \sqrt{z_{1}^{2}+\left(948-x_{1}-\cdots-x_{n}\right)^{2}+\left(1185-y_{1}-\cdots-y_{n}\right)^{2}}
\end{aligned}
$$

where $x_{i}, y_{i}, z_{i}, i=1,2, \cdots, n$, are non-negative real numbers.
[Solution] Consider the following points in a three-dimensional space with a Cartesian coordinate system:
$$
\begin{array}{l}
B=(0,0,1264) ; \\
M_{n}=\left(x_{n}, y_{n}, z_{1}+\cdots+z_{n}\right), \\
M_{n-1}=\left(x_{n-1}+x_{n}, y_{n-1}+y_{n}, z_{1}+\cdots+z_{n-1}\right), \\
\cdots \cdots \cdots \cdots \\
M_{2}=\left(x_{2}+\cdots+x_{n}, y_{2}+\cdots+y_{n}, z_{1}+z_{2}\right), \\
M_{1}=\left(x_{1}+\cdots+x_{n}, y_{1}+\cdots y_{n}, z_{1}\right), \\
C=(948,1185,0),
\end{array}
$$

Clearly, \( A = B M_{n} + M_{n} M_{n-1} + \cdots + M_{2} M_{1} + M_{1} C \). By the triangle inequality, \( A \geqslant B C = \sqrt{948^2 + 1185^2 + 1264^2} = 1975 \). On the other hand, if we take \( M_{1} = M_{2} = \cdots = M_{n} = B \), then \( A = B C = 1975 \). Therefore, the minimum value of \( A \) is 1975.