In $∆\mathrm{ABC}$, L,M,N are points on BC, CA, AB respectively, dividing them in the ratio 1:2, 2 :  3, 3 : 5. If the point K divides AB in the ratio 5:3, then $\frac{\left|\overline{\mathrm{AL}}+\overline{\mathrm{BM}}+\overline{\mathrm{CN}}\right|}{\left|\overline{\mathrm{CK}}\right|}=$

$$\begin{gathered} \overline{\mathrm{OA}}=\overline{\mathrm{a}} \overline{\mathrm{OB}}=\overline{\mathrm{b}} \overline{ \mathrm{OC}}=\overline{\mathrm{c}} \\ \overline{\mathrm{OL}}=\frac{\overline{\mathrm{c}}+2\overline{\mathbf{b}}}{3} \overline{\mathrm{OM}}=\frac{3\overline{\mathrm{c}}}{5} \overline{\mathrm{ON}}=\frac{3\overline{\mathrm{b}}}{8} \\ \overline{\mathrm{OK}}=\frac{5\overline{\mathrm{b}}}{8} \\ \left|\overline{\mathrm{AL}}+\overline{\mathrm{BM}}+\overline{\mathrm{CN}}\right|=\left|\frac{\overline{\mathrm{c}}+2\overline{\mathrm{b}}}{3}+\frac{3\overline{\mathrm{c}}}{5}-\overline{\mathrm{b}}+\frac{3\overline{\mathrm{b}}}{8}-\overline{\mathrm{c}}\right| \\ =\left|\frac{40\overline{\mathrm{c}}+80\overline{\mathrm{b}}+72\overline{\mathrm{c}}-120\overline{\mathrm{b}}+45\overline{\mathrm{b}}-120\overline{\mathrm{c}}}{120}\right| \\ \left|\overline{\mathrm{CK}}\right|=\left|\frac{5\overline{\mathrm{b}}}{8}-\overline{\mathrm{c}}\right|=\frac{\left|5\overline{\mathrm{b}}-8\overline{\mathrm{c}}\right|}{8} \\ =\frac{{\displaystyle \frac{1}{120}}}{{\displaystyle \frac{1}{8}}}=\frac{1}{15} \end{gathered}$$