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

The line is given in symmetric form
$$\frac{x + 1}{1}=\frac{y - 1}{2}=\frac{z - 2}{2}$$
so its direction ratios (d.r.’s) are $$1,\,2,\,2$$. Hence its direction vector can be taken as $$\mathbf{d}=\langle 1,\,2,\,2\rangle$$ with magnitude
$$|\mathbf{d}|=\sqrt{1^{2}+2^{2}+2^{2}}=\sqrt{9}=3.$$

The plane is
$$2x-y+\sqrt{\lambda}\,z+4=0,$$
so a normal vector to the plane is $$\mathbf{n}=\langle 2,\,-1,\,\sqrt{\lambda}\rangle$$ with magnitude
$$|\mathbf{n}|=\sqrt{2^{2}+(-1)^{2}+(\sqrt{\lambda})^{2}}=\sqrt{4+1+\lambda}=\sqrt{5+\lambda}.$$

Let $$\theta$$ be the acute angle between the line and the plane. If $$\phi$$ is the angle between the line and the normal to the plane, then $$\theta+\phi=90^{\circ}$$, so $$\sin\theta=\cos\phi.$$

Using the dot‐product formula, $$\cos\phi=\frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}|\,|\mathbf{n}|}.$$ Hence $$\sin\theta=\frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}|\,|\mathbf{n}|}.$$

Compute the dot product: $$\mathbf{d}\cdot\mathbf{n}=1\cdot2+2\cdot(-1)+2\cdot\sqrt{\lambda}=2-2+2\sqrt{\lambda}=2\sqrt{\lambda}.$$ Because $$\sqrt{\lambda}\ge 0$$ for real $$\lambda$$, the absolute value is $$|\mathbf{d}\cdot\mathbf{n}|=2\sqrt{\lambda}.$$

Given $$\sin\theta=\frac13,$$ we get $$\frac{2\sqrt{\lambda}}{3\sqrt{5+\lambda}}=\frac13.$$

Cross-multiply and simplify: $$2\sqrt{\lambda}=\,\sqrt{5+\lambda}.$$ Square both sides: $$4\lambda=5+\lambda\quad\Longrightarrow\quad 3\lambda=5\quad\Longrightarrow\quad \lambda=\frac53.$$

Thus the required value of $$\lambda$$ is $$\frac53$$.

Option A which is: $$\frac{5}{3}$$