Show that the three lines with direction cosines $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13} ; \frac{4}{13}, \frac{12}{13}, \frac{3}{13} ; \frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ are mutually perpendicular.

Two lines with direction cosines $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ are perpendicular to each other if $l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} = 0$.
$(i)$ For the lines with direction cosines $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$ and $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$,we have:
$l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} = \left(\frac{12}{13}\right)\left(\frac{4}{13}\right) + \left(\frac{-3}{13}\right)\left(\frac{12}{13}\right) + \left(\frac{-4}{13}\right)\left(\frac{3}{13}\right)$
$= \frac{48}{169} - \frac{36}{169} - \frac{12}{169} = 0$.
Thus,these two lines are perpendicular.
$(ii)$ For the lines with direction cosines $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$ and $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$,we have:
$l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} = \left(\frac{4}{13}\right)\left(\frac{3}{13}\right) + \left(\frac{12}{13}\right)\left(\frac{-4}{13}\right) + \left(\frac{3}{13}\right)\left(\frac{12}{13}\right)$
$= \frac{12}{169} - \frac{48}{169} + \frac{36}{169} = 0$.
Thus,these two lines are perpendicular.
$(iii)$ For the lines with direction cosines $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ and $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$,we have:
$l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} = \left(\frac{3}{13}\right)\left(\frac{12}{13}\right) + \left(\frac{-4}{13}\right)\left(\frac{-3}{13}\right) + \left(\frac{12}{13}\right)\left(\frac{-4}{13}\right)$
$= \frac{36}{169} + \frac{12}{169} - \frac{48}{169} = 0$.
Thus,these two lines are perpendicular.
Since all pairs are perpendicular,the three lines are mutually perpendicular.