The given lines are $$2x = 3y = - z$$

or $$\,\,\,\,\,\,\,\,\,\,{x \over 3} = {y \over 2} = {z \over { - 6}}$$ $$\,\,\,\left[ {} \right.$$ Dividing by $$6$$ $$\left. {} \right]$$

and $$6x = - y = - 4z$$

or $$\,\,\,\,\,\,\,\,\,{x \over 2} = {y \over { - 12}} = {z \over { - 3}}$$ $$\,\,\,\,\left[ {} \right.$$ Dividing by $$12$$ $$\left. {} \right]$$

$$\therefore$$ Angle between two lines is

$$\cos \theta = {{3.2 + 2.\left( { - 12} \right) + \left( { - 6} \right).\left( { - 3} \right)} \over {\sqrt {{3^2} + {2^2} + {{\left( { - 6} \right)}^2}} \sqrt {{2^2} + {{\left( { - 12} \right)}^2} + {{\left( { - 3} \right)}^2}} }}$$

$$ = {{6 - 24 + 18} \over {\sqrt {49} \sqrt {157} }} = 0 \Rightarrow \theta = {90^ \circ }$$