Given lines can be written as
$\frac{x}{\left(\frac{1}{2}\right)}=\frac{y}{\left(\frac{1}{3}\right)}=\frac{z}{-1}$
and $\frac{x}{\left(\frac{1}{6}\right)}=\frac{y}{(-1)}=\frac{z}{\left(-\frac{1}{4}\right)}$
If $\theta$ is angle between them
$\cos \theta=\frac{\frac{1}{2} \times \frac{1}{6}+\frac{1}{3}(-1)+(-1) \times\left(-\frac{1}{4}\right)}{\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{3}\right)^{2}+(-1)^{2}}}$
$ \sqrt{\left(\frac{1}{6}\right)^{2}+(-1)^{2}+\left(-\frac{1}{4}\right)^{2}} $
$=0 \Rightarrow \theta=\frac{\pi}{2} $