Given that, in $\triangle A B C, L, M, N$ are points on BC, $\mathrm{CA}$ and $\mathrm{AB}$ respectively, dividing in the ratio $1: 2,2: 3$ and $3: 5$. Also, point $K$ divides $A B$ in the ratio $5: 3$.


Now, according to the section formula,
$$
\begin{aligned}
& L=\frac{2 \mathbf{b}+\mathbf{c}}{3}, M=\frac{2 \mathbf{a}+3 \mathbf{c}}{5} \\
& N=\frac{3 \mathbf{b}+5 \mathbf{a}}{8}, K=\frac{5 \mathbf{b}+3 \mathbf{a}}{8} \\
& \overrightarrow{\mathrm{AL}}=\frac{2 \mathbf{b}+\mathbf{c}}{3}-\mathbf{a}=\frac{2 \mathbf{b}+\mathbf{c}-3 \mathbf{a}}{3} \\
& \overrightarrow{\mathrm{BM}}=\frac{2 \mathbf{a}+3 \mathbf{c}}{5}-\mathbf{b}=\frac{2 \mathbf{b}+3 \mathbf{c}-5 \mathbf{b}}{5} \\
& \overrightarrow{\mathrm{CN}}=\frac{3 \mathbf{b}+5 \mathbf{a}}{8}-\mathbf{c}=\frac{3 \mathbf{b}+5 \mathbf{a}-8 \mathbf{c}}{8} \\
& \overrightarrow{\mathrm{CK}}=\frac{5 \mathbf{b}+3 \mathbf{a}}{8}-\mathbf{c}=\frac{5 \mathbf{b}+3 \mathbf{a}-8 \mathbf{c}}{8} \\
& \because \quad \frac{\overrightarrow{\mathrm{AL}}+\overrightarrow{\mathrm{BM}}+\overrightarrow{\mathrm{CN}}}{\overrightarrow{\mathrm{CK}}} \mid=
\end{aligned}
$$
$\begin{aligned} & \left|\frac{\frac{2 \mathbf{b}+\mathbf{c}-3 \mathbf{a}}{3}+\frac{2 \mathbf{a}+3 \mathbf{c}-5 \mathbf{b}}{5}+\frac{3 \mathbf{b}+5 \mathbf{a}-8 \mathbf{c}}{8}}{\frac{5 \mathbf{b}+3 \mathbf{a}-8 \mathbf{c}}{8}}\right| \\ = & \left|\frac{(5 \mathbf{b}+3 \mathbf{a}-8 \mathbf{c})}{15(5 \mathbf{b}+3 \mathbf{a}-8 \mathbf{c})}\right|=\frac{1}{15}\end{aligned}$