1. Let the direction cosines be $l,m,n$.

   They satisfy: $2l+2m-n=0$ $mn+nl+lm=0$ and since they are direction cosines, $l^2+m^2+n^2=1$

2. Use the linear relation to express $n$: $n=2l+2m$

3. Substitute into $mn+nl+lm=0$: $m(2l+2m)+l(2l+2m)+lm=0$ Expanding, $2lm+2m^2+2l^2+2lm+lm=0$ $2l^2+5lm+2m^2=0$

4. Solve for the ratio $\dfrac{l}{m}$.

   Let $x=\dfrac{l}{m}$. Then $2x^2+5x+2=0$ $(2x+1)(x+2)=0$ So, $x=-\frac12 \quad \text{or} \quad x=-2$

   Hence the two possible directions are obtained from:

   • $l:m=-1:2$
   • $l:m=-2:1$

5. Find corresponding $n$ using $n=2l+2m$.

   • If $l:m=-1:2$, then $n=2(-1)+2(2)=2$ so direction ratios are proportional to $(-1,2,2)$

   • If $l:m=-2:1$, then $n=2(-2)+2(1)=-2$ so direction ratios are proportional to $(-2,1,-2)$

6. Find the angle between the two lines.

   If the direction ratios are $\vec a=(-1,2,2), \quad \vec b=(-2,1,-2),$ then $\vec a\cdot \vec b = (-1)(-2)+(2)(1)+(2)(-2)=2+2-4=0$

   Therefore, $\cos\theta=0 \implies \theta=\frac{\pi}{2}$

7. Check options:

   • A: $\dfrac{\pi}{2}$ ✅
   • B, C, D are incorrect.

Therefore, the angle between the lines is $\boxed{\frac{\pi}{2}}$