If the angle $\theta$ between the line $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2x-y+\lambda z+4=0$ is such that $\sin \theta =1/3$, then the value of $\lambda$ is:

The direction ratios of the line are $\vec{b}=(1,2,2)$.The normal vector to the plane is $\vec{n}=(2,-1,\lambda )$.The angle $\theta$ between a line and a plane is given by:
$$\sin \theta =\frac{|\vec{b}\cdot \vec{n}|}{|\vec{b}||\vec{n}|}$$
Given $\sin \theta =1/3$:
$$\frac{1}{3}=\frac{|(1)(2)+(2)(-1)+(2)(\lambda )|}{\sqrt{1^2+2^2+2^2}\sqrt{2^2+(-1{)}^2+{\lambda }^2}}$$
$$\frac{1}{3}=\frac{|2-2+2\lambda |}{\sqrt{9}\sqrt{5+{\lambda }^2}}$$
$$\frac{1}{3}=\frac{|2\lambda |}{3\sqrt{5+{\lambda }^2}}$$
$$1=\frac{|2\lambda |}{\sqrt{5+{\lambda }^2}}$$
Squaring both sides:
$$1=\frac{4{\lambda }^2}{5+{\lambda }^2}$$
$$5+{\lambda }^2=4{\lambda }^2⟹3{\lambda }^2=5⟹{\lambda }^2=5/3$$
Wait, checking calculation: $|2\lambda |/\sqrt{5+{\lambda }^2}=1⟹5+{\lambda }^2=4{\lambda }^2$. Yes.Let's check the provided source answer (1) which is $5/3$. If $\lambda =5/3$, ${\lambda }^2=25/9\ne 5/3$.Let's re-read the normal: $2x-y+\lambda z+4=0⟹\vec{n}=(2,-1,\lambda )$.The numerator is $2-2+2\lambda =2\lambda$.If the correct value is $\lambda =5/3$ as per the key, let's work backward.$\frac{2(5/3)}{\sqrt{5+25/9}}=\frac{10/3}{\sqrt{70/9}}\ne 1$.There might be a typo in the question's direction ratios or plane eq. However, following the math leads to $\lambda =\pm \sqrt{5/3}$.