The angle between the lines $2x=3y=-z$ and $6x=-y=-4z$ is:

First, find the direction ratios of the lines.Line 1: $2x=3y=-z⟹\frac{x}{1/2}=\frac{y}{1/3}=\frac{z}{-1}$ Multiply by 6: $\frac{x}{3}=\frac{y}{2}=\frac{z}{-6}$ So, $\overrightarrow{d_1}=(3,2,-6)$.Line 2: $6x=-y=-4z⟹\frac{x}{1/6}=\frac{y}{-1}=\frac{z}{-1/4}$ Multiply by 12: $\frac{x}{2}=\frac{y}{-12}=\frac{z}{-3}$ So, $\overrightarrow{d_2}=(2,-12,-3)$.Check the dot product $\overrightarrow{d_1}\cdot \overrightarrow{d_2}$:$3(2)+2(-12)+(-6)(-3)=6-24+18=0$.Since the dot product of the direction ratios is zero, the lines are perpendicular.The angle is $90^{\circ }$.