If the angle θ between the line $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2x - y + \sqrt{λz}+ 4 = 0 $ i such that $sin θ = \frac{1}{3}$, the value of λ is

$\frac{5}{3}$

The given line is parallel to the vector $\vec{b} = \hat{i} +2\hat{j} + 2\hat{k}$ and the given plane is normal to the vector $\vec{n} = 2\hat{i} - \hat{j} + \sqrt{λ}\hat{k}.$

$∴ sin θ =\frac{\vec{b}.\vec{n}}{|\vec{b}||\vec{n}|}$

$⇒ \frac{1}{3}=\frac{2-2+2\sqrt{λ}}{3×\sqrt{λ+5}}⇒\sqrt{λ+5} =2\sqrt{λ}⇒λ =\frac{5}{3}$