$\mathrm{n}_{1}: \mathrm{n}_{2}: \mathrm{n}_{3}=1: 3: 4$

Let $L_{1}, L_{2}$ and $L_{3}$ be the lengths of the three parts

As $n \propto \frac{1}{L}$

$\therefore \quad \mathrm{L}_{1}: \mathrm{L}_{2}: \mathrm{L}_{3}=\frac{1}{1}: \frac{1}{3}: \frac{1}{4}=12: 4: 3$

$\therefore \quad \mathrm{L}_{1}=72 \mathrm{cm}\left(\frac{12}{12+4+3} \times 114\right)$

$\mathrm{L}_{2}=24 \mathrm{cm}\left(\frac{4}{19} \times 114\right)$

and $\mathrm{L}_{3}=18 \mathrm{cm}\left(\frac{3}{19} \times 114\right)$

Hence the bridges should be placed at

$72 \mathrm{cm}$ and $72+24=96 \mathrm{cm}$ from oneend