Wired

The least length of power line is obtained by joining all three cities $A,B,C$ to the Fermat point $P$ of the triangle $ABC$ . This point can be found by constructing equilateral triangles on the sides of the triangle $ABC$ , and then joining the new vertices of each equilateral triangle to the opposite vertex of $ABC$ . These three lines are concurrent at $P$ . In the diagram we have constructed the equilateral triangles $ABD$ and $BCE$ , and the lines $DC$ and $EA$ intersect at $P$ .

The least length of power lines is the length $AP + BP + CP = DC$ . Now $\begin{aligned} \cos \angle BAC & = \; \frac{13^2 + 21^2 - 17^2}{2\times13\times21} = \frac{107}{182} \\ \sin \angle BAC & = \; \frac{85\sqrt{3}}{182} \\ \cos \angle DAC & = \; \cos(\angle BAC + 60^\circ) \; = \; \frac{107}{182} \times \frac12 - \frac{85\sqrt{3}}{182} \times \frac{\sqrt{3}}{2} \; = \; -\frac{37}{91} \end{aligned}$ and hence $DC^2 \; = \; 13^2 + 21^2 - 2 \times 13 \times 21 \times \cos \angle DAC \;= \; 832$ making the least length of power line $\boxed{8\sqrt{13}}$ miles.

Two word hint.........Fermat Point.........!!!
Also, This note might be helpful..........!!