Let the direction cosines of the three lines be denoted as $l_1, m_1, n_1$, $l_2, m_2, n_2$, and $l_3, m_3, n_3$ respectively. From the provided context, two lines with direction cosines $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are perpendicular if $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.

For lines 1 and 2, the sum of products is: $(\frac{12}{13})(\frac{4}{13}) + (\frac{-3}{13})(\frac{12}{13}) + (\frac{-4}{13})(\frac{3}{13}) = \frac{48}{169} - \frac{36}{169} - \frac{12}{169} = 0$. Hence, Line 1 is perpendicular to Line 2. For lines 2 and 3, the sum of products is: $(\frac{4}{13})(\frac{3}{13}) + (\frac{12}{13})(\frac{-4}{13}) + (\frac{3}{13})(\frac{12}{13}) = \frac{12}{169} - \frac{48}{169} + \frac{36}{169} = 0$. Hence, Line 2 is perpendicular to Line 3. For lines 1 and 3, the sum of products is: $(\frac{12}{13})(\frac{3}{13}) + (\frac{-3}{13})(\frac{-4}{13}) + (\frac{-4}{13})(\frac{12}{13}) = \frac{36}{169} + \frac{12}{169} - \frac{48}{169} = 0$. Hence, Line 1 is perpendicular to Line 3. Since each pair is perpendicular, the lines are mutually perpendicular.