11. Let $w_{1}, w_{4}, \cdots, m_{n}$ be complex numbers. If a straight line $l$ passes through points (complex numbers) $z_{1}: z_{2}, \cdots, z_{n}$, such that $\sum_{k=1}^{n}\left(z_{k}-w_{k}\right)=0$, then $l$ is called the “average line” of $w_{1}, w_{i}$, $\cdots, ~ w w_{\mathrm{n}}$.

For $w_{1}=32+170 i, w_{2}=-7+64 i$, $w_{3}=-9+200 i, w_{4}=1+27 i, w_{5}=-14$ $+43 i$, there is a unique “average line”, whose y-intercept is $y=3$. Find the slope of this line.
11. (163)

Let $y=m x+b$ be the "average line" of the complex numbers $w_{\mathrm{K}}=u_{\mathrm{K}}+i v_{\mathrm{K}}$, where $u_{\mathrm{K}}, v_{\mathrm{K}} \in R, k=1,2, \cdots, n$. Assume that on the line $y=m x+b$, we select complex numbers $z_{\mathrm{K}}=x_{\mathrm{K}}+i y_{\mathrm{K}}\left(x_{\mathrm{K}}, y_{\mathrm{K}} \in R, 1 \leqslant k \leqslant n\right)$ such that
$$
\begin{array}{c}
\sum_{\mathrm{k}=1}^{\mathrm{n}}\left(z_{\mathrm{K}}-w_{\mathrm{K}}\right)=0 . \\
\text { Then } \sum x_{\mathrm{K}}=\sum u_{\mathrm{K}}, \Sigma y_{\mathrm{K}}=\Sigma v_{\mathrm{K}}, \text { and } y_{\mathrm{K}}=m x_{\mathrm{K}}+ \\
b(1 \leqslant k \leqslant n) . \text { Therefore, } \Sigma v_{\mathrm{K}}=\Sigma y_{\mathrm{K}}=\Sigma\left(m x_{\mathrm{K}}\right. \\
+b)=m \Sigma x_{\mathrm{K}}+n b=\left(\Sigma u_{\mathrm{K}}\right) m+n b . \\
\text { The problem gives } n=5, b=3, \Sigma u_{\mathrm{K}}=32 \\
+(-7)+(-9)+1+(-14)=3, \\
\sum v_{\mathrm{K}}=170+64+200+27+43=504 . \\
\therefore \quad 504=3 m+5 \text {, so } m=163 .
\end{array}
$$