Solution

Let the angle between the lines be $$\alpha$$

Given Direction cosines of two lines are $$$$

Also given that $$2 l+2 m-n=0\implies n=2 l+2 m\cdots(1)$$

$$m n+n l+l m=0\cdots(2)$$

$$\implies m(2 l+2 m)+l(2 l+2 m)+l m=0$$ $$(\because \text{from}(1)\ )$$

$$\implies 2 m^2+5 l m+2 l^2=0$$

$$\implies 2 m^2+4 l m+l m+2 l^2=0$$

$$\implies 2 m(m+2 l)+l(m+2 l)=0$$

$$\implies (2 m+l)(m+2 l)=0$$

$$\implies m=-\dfrac{l}{2},-2 l$$

$$n=2 l+2 m=2 l+2\bigg(-\dfrac{l}{2}\bigg),2 l+2(-2 l)=2 l-l,2 l-4 l=l,-2 l$$

So the direction cosines of two lines are $$$$ and $$$$

So the direction ratios will be $$<2,-1,2>$$ and $$<1,-2,-2>$$

As we know that

If $$\theta$$ is the angle between the two lines having direction ratios as $$$$ and $$$$
then $$\cos \theta=\dfrac{a d+be +c f}{\sqrt{a^2+b^2+c^2}\sqrt{d^2+e^2+f^2}}$$

Hence $$\cos \alpha=\dfrac{(2)(1)+(-1)(-2)+2(-2)}{\sqrt{2^2+(-1)^2+2^2}\sqrt{1^2+(-2)^2+(-2)^2}}=\dfrac{2+2-4}{3\times 3}=0$$

$$\implies \alpha=\dfrac{\pi}{2}$$

So the angle between the two lines is $$\dfrac{\pi}{2}$$