1. **State the problem:** Show that the three lines with direction cosines $$\left(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\right), \quad \left(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\right), \quad \left(\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\right)$$ are mutually perpendicular. 2. **Recall the rule for perpendicularity:** Two lines are perpendicular if their direction vectors have a dot product of zero. 3. **Calculate dot products:** Let vectors be $$\mathbf{a} = \left(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\right), \quad \mathbf{b} = \left(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\right), \quad \mathbf{c} = \left(\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\right)$$ Calculate $\mathbf{a} \cdot \mathbf{b}$: $$\frac{12}{13} \times \frac{4}{13} + \frac{-3}{13} \times \frac{12}{13} + \frac{-4}{13} \times \frac{3}{13} = \frac{48}{169} - \frac{36}{169} - \frac{12}{169} = \frac{48 - 36 - 12}{169} = \frac{0}{169} = 0$$ Calculate $\mathbf{b} \cdot \mathbf{c}$: $$\frac{4}{13} \times \frac{3}{13} + \frac{12}{13} \times \frac{-4}{13} + \frac{3}{13} \times \frac{12}{13} = \frac{12}{169} - \frac{48}{169} + \frac{36}{169} = \frac{12 - 48 + 36}{169} = \frac{0}{169} = 0$$ Calculate $\mathbf{c} \cdot \mathbf{a}$: $$\frac{3}{13} \times \frac{12}{13} + \frac{-4}{13} \times \frac{-3}{13} + \frac{12}{13} \times \frac{-4}{13} = \frac{36}{169} + \frac{12}{169} - \frac{48}{169} = \frac{36 + 12 - 48}{169} = \frac{0}{169} = 0$$ 4. **Conclusion:** Since all dot products are zero, the three lines are mutually perpendicular.