Step 1: Understanding the Concept:
The angle $\theta$ between a line and a plane is the complement of the angle between the line's direction vector and the plane's normal vector. Therefore, the sine of the angle $\theta$ between them is given by the cosine formula for the two vectors.
Step 2: Key Formula or Approach:
If a line has direction vector $\vec{b} = \langle b_1, b_2, b_3 \rangle$ and a plane has normal vector $\vec{n} = \langle n_1, n_2, n_3 \rangle$, the angle $\theta$ between them satisfies:
\[ \sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|} \]
Step 3: Detailed Explanation:
From the line equation $\frac{x+1}{1} = \frac{y-1}{2} = \frac{z-2}{2}$, the direction vector is:
\[ \vec{b} = \langle 1, 2, 2 \rangle \]
From the plane equation $2x - y + \sqrt{\lambda}z + 4 = 0$, the normal vector is:
\[ \vec{n} = \langle 2, -1, \sqrt{\lambda} \rangle \]
Calculate the dot product $\vec{b} \cdot \vec{n}$:
\[ \vec{b} \cdot \vec{n} = (1)(2) + (2)(-1) + (2)(\sqrt{\lambda}) = 2 - 2 + 2\sqrt{\lambda} = 2\sqrt{\lambda} \]
Calculate the magnitudes of the vectors:
\[ |\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]
\[ |\vec{n}| = \sqrt{2^2 + (-1)^2 + (\sqrt{\lambda})^2} = \sqrt{4 + 1 + \lambda} = \sqrt{5 + \lambda} \]
Now apply the angle formula, given $\sin \theta = \frac{1}{3}$:
\[ \frac{1}{3} = \frac{|2\sqrt{\lambda}|}{3\sqrt{5 + \lambda}} \]
Multiply both sides by 3:
\[ 1 = \frac{2\sqrt{\lambda}}{\sqrt{5 + \lambda}} \]
Since $\lambda$ must be positive for the square root to be real in this context, we can drop the absolute value. Square both sides to eliminate the square roots:
\[ 1^2 = \left( \frac{2\sqrt{\lambda}}{\sqrt{5 + \lambda}} \right)^2 \]
\[ 1 = \frac{4\lambda}{5 + \lambda} \]
Multiply by $(5 + \lambda)$:
\[ 5 + \lambda = 4\lambda \]
\[ 5 = 3\lambda \implies \lambda = \frac{5}{3} \]
We are asked to find the value of $\lambda + 1$:
\[ \lambda + 1 = \frac{5}{3} + 1 = \frac{5 + 3}{3} = \frac{8}{3} \]
Step 4: Final Answer:
The value of $\lambda + 1$ is $\frac{8}{3}$.