Given line is $L:\frac{x+1}{1}=\frac{y}{2}=\frac{z+4}{2}$ and plane is $P\mathrm{:} 2x-y+\sqrt{\lambda }z+4=0.$

The angle between a line $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$ and a plane $ax+by+cz+d=0$ is 

$\sin \theta =\frac{\left|al+bm+cn\right|}{\sqrt{a^2+b^2+c^2}\sqrt{l^2+m^2+n^2}}$


         

$\sin \left(\frac{\mathrm{\pi }}{6}\right)=\frac{\left|\left(2\times 1\right)+\left(-1\times 2\right)+2\sqrt{\lambda }\right|}{\sqrt{1^2+2^2+2^2}\sqrt{2^2+1^2+\lambda }}$

$\Rightarrow \frac{1}{2}=\frac{\left|2-2+2\sqrt{\lambda }\right|}{\sqrt{1+4+4}\sqrt{4+1+\lambda }}$

$\Rightarrow \frac{1}{2}=\frac{\left|2\sqrt{\lambda }\right|}{3\sqrt{5+\lambda }}$

$\Rightarrow 4\sqrt{\lambda }=3\sqrt{5+\lambda }$

Squaring both sides, we get

$\Rightarrow 16\lambda =9\left(5+\lambda \right)$

$\Rightarrow 16\lambda =45+9\lambda$

$\Rightarrow 7\lambda =45$

$\Rightarrow \lambda =\frac{45}{7}.$