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

The given equations of the lines are $2x = 3y = -z$ and $6x = -y = -4z$.

Rewrite the first line in standard symmetric form $\dfrac{x}{a_1} = \dfrac{y}{b_1} = \dfrac{z}{c_1}$. Dividing by the least common multiple of the coefficients (which is 6), we get $\dfrac{x}{3} = \dfrac{y}{2} = \dfrac{z}{-6}$. Thus, the direction vector of the first line is $\vec{v_1} = 3\hat{i} + 2\hat{j} - 6\hat{k}$.

Rewrite the second line in standard symmetric form $\dfrac{x}{a_2} = \dfrac{y}{b_2} = \dfrac{z}{c_2}$. Dividing by the least common multiple of the coefficients (which is 12), we get $\dfrac{x}{2} = \dfrac{y}{-12} = \dfrac{z}{-3}$. Thus, the direction vector of the second line is $\vec{v_2} = 2\hat{i} - 12\hat{j} - 3\hat{k}$.

The cosine of the angle $\theta$ between the two lines is given by $\cos \theta = \dfrac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.

Calculate the dot product: $\vec{v_1} \cdot \vec{v_2} = (3)(2) + (2)(-12) + (-6)(-3) = 6 - 24 + 18 = 0$.

Since the dot product is $0$, the vectors are perpendicular, which means $\theta = 90^{\circ}$.

Answer: $90^{\circ}$