1. Find direction ratios of each line

For the line $2x=3y=-z$ let the common value be $\lambda$. Then $2x=\lambda,\quad 3y=\lambda,\quad -z=\lambda$ so $x=\frac{\lambda}{2},\quad y=\frac{\lambda}{3},\quad z=-\lambda$ Hence a direction ratio vector is $\left(\frac12,\frac13,-1\right)$ Multiplying by $6$, we can take direction ratios as $\vec d_1=(3,2,-6).$

For the line $6x=-y=-4z$ let the common value be $\mu$. Then $6x=\mu,\quad -y=\mu,\quad -4z=\mu$ so $x=\frac{\mu}{6},\quad y=-\mu,\quad z=-\frac{\mu}{4}$ Hence a direction ratio vector is $\left(\frac16,-1,-\frac14\right)$ Multiplying by $12$, we can take direction ratios as $\vec d_2=(2,-12,-3).$

2. Use the angle formula between two lines

If $\theta$ is the angle between the lines, then $\cos\theta=\frac{|\vec d_1\cdot \vec d_2|}{|\vec d_1|\,|\vec d_2|}.$

Compute the dot product: $\vec d_1\cdot \vec d_2=(3)(2)+(2)(-12)+(-6)(-3)=6-24+18=0.$

Thus, $\cos\theta=0.$ So, $\theta=90^\circ.$

3. Check options

• A: $0^\circ$ ❌
• B: $90^\circ$ ✅
• C: $45^\circ$ ❌
• D: $30^\circ$ ❌

Therefore, the angle between the lines is $\boxed{90^\circ}.$