The lines $$2x = 3y = - z$$    and $$6x = - y = - 4z$$

Solution :
$$\eqalign{ & 2x = 3y = - z \cr & {\text{or }}\frac{x}{3} = \frac{y}{2} = \frac{z}{{ - 6}} \cr & 6x = - y = - 4z \cr & {\text{or }}\frac{x}{2} = \frac{y}{{ - 12}} = \frac{z}{{ - 3}} \cr & \cos \,\theta = \frac{{{x_1}{x_2} + {y_1}{y_2} + {z_1}{z_2}}}{{\sqrt {x_1^2 + x_2^2 + x_3^2} .\sqrt {y_1^2 + y_2^2 + y_3^2} }} \cr & \cos \,\theta = \frac{{\left( {6 - 24 + 18} \right)}}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( { - 6} \right)}^2}} .\sqrt {{{\left( 2 \right)}^2} + {{\left( { - 12} \right)}^2} + {{\left( { - 3} \right)}^2}} }} \cr & \cos \,\theta = 0 \cr & \theta = {90^ \circ } \cr} $$
So lines are perpendicular.