Solution:

Consider the direction cosines of $L_{1}, L_{2}$ and $L_{3}$ be $l_{1}, m_{1}, n_{1} ; l_{2}, m_{2}, n_{2}$ and $l_{3}, m_{3}, n_{3}$.
It is known that
If $\mathrm{f}_{1}, \mathrm{~m}_{1}, \mathrm{n}_{1}$ and $\mathrm{l}_{2}, \mathrm{~m}_{2}, \mathrm{n}_{2}$ are the direction cosines of two lines;
And $\theta$ is the acute angle between the two lines;
Therefore $\cos \theta=\|_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2} \mid$
If two lines are perpendicular, then the angle between the two is $\theta=90^{\circ}$
For perpendicular lines, $\left|I_{1}\right|_{2}+m_{1} m_{2}+n_{1} n_{2} \mid=\cos 90^{\circ}=0$, i.e. $\left|I_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}\right|=0$
Therefore, in order to check if the three lines are mutually perpendicular, we compute $\left|I_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}\right|$ for all the pairs of the three lines.
Firstly let’s compute, $\left|I_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}\right|$
$\begin{array}{l}
\left|1_{1} 1_{2}+m_{1} m_{2}+n_{1} n_{2}\right|=\left|\left(\frac{12}{13} \times \frac{4}{13}\right)+\left(\frac{-3}{13} \times \frac{12}{13}\right)+\left(\frac{-4}{13} \times \frac{3}{13}\right)\right|=\frac{48}{13}+\left(\frac{-36}{13}\right)+\left(\frac{-12}{13}\right) \\
=\frac{48+(-48)}{13}=0
\end{array}$
Therefore, $L_{1} \perp L_{2} \ldots \ldots$ (1)
In the similar way,
Let’s compute, $\mid \mathrm{l}_{2} 1_{3}+\mathrm{m}_{2} \mathrm{~m}_{3}+\mathrm{n}_{2} \mathrm{n}_{3} \mathrm{l}$
$\begin{array}{l}
\left|1_{2} 1_{3}+m_{2} m_{3}+n_{2} n_{3}\right|=\left|\left(\frac{4}{13} \times \frac{3}{13}\right)+\left(\frac{12}{13} \times \frac{-4}{13}\right)+\left(\frac{3}{13} \times \frac{12}{13}\right)\right|=\frac{12}{13}+\left(\frac{-48}{13}\right)+\frac{36}{13} \\
=\frac{(-48)+48}{13}=0
\end{array}$
So, $L_{2} \perp L_{3} \ldots . .(2)$
In the similar way,
Let’s compute, $\left|\mathrm{I}_{3}\right|_{1}+\mathrm{m}_{3} \mathrm{~m}_{1}+\mathrm{n}_{3} \mathrm{n}_{1} \mathrm{l}$
$\begin{array}{l}
\left|1_{3} 1_{1}+\mathrm{m}_{3} \mathrm{~m}_{1}+\mathrm{n}_{3} \mathrm{n}_{1}\right|=\left|\left(\frac{3}{13} \times \frac{12}{13}\right)+\left(\frac{-4}{13} \times \frac{-3}{13}\right)+\left(\frac{12}{13} \times \frac{-4}{13}\right)\right|=\frac{36}{13}+\frac{12}{13}+\left(\frac{-48}{13}\right) \\
=\frac{48+(-48)}{13}=0
\end{array}$
So, $L_{1} \perp L_{3} \ldots . .(3)$
As a result, by $(1),(2)$ and (3), the lines are perpendicular.
$\mathrm{L}_{1}, \mathrm{~L}_{2}$ and $\mathrm{L}_{3}$ are mutually perpendicular.