If the angle `theta` between the line `(x+1)/1=(y-1)/2=(z-2)/2` and the plane `2x-y+sqrt(pz)+4=0` is such that `sintheta=1/3`, then the values of p is (A) 0 (B) `1/3` (C) `2/3` (D) `5/3`

Options

A. `(-3)/(5)`
B. `(5)/(3)`
C. `(-4)/(3)`
D. `(3)/(4)`

Explanation

To solve the problem, we need to find the value of \( p \) given the angle \( \theta \) between the line and the plane. The sine of the angle is given as \( \sin \theta = \frac{1}{3} \).

### Step 1: Identify the direction ratios of the line and the normal to the plane.

The line is given by the equation:
\[ \frac{x + 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{2} \]
From this, we can extract the direction ratios of the line:
\[ \vec{l_1} = (1, 2, 2) \]

The plane is given by the equation:
\[ 2x - y + \sqrt{p}z + 4 = 0 \]
The normal vector to the plane can be identified as:
\[ \vec{n} = (2, -1, \sqrt{p}) \]

### Step 2: Use the relationship between the angle and the sine.

The angle \( \theta \) between the line and the plane is related to the angle \( \theta' \) between the line and the normal to the plane:
\[ \sin \theta = \cos \theta' \]
Given that \( \sin \theta = \frac{1}{3} \), we have:
\[ \cos \theta' = \frac{1}{3} \]

### Step 3: Calculate the dot product of the line's direction vector and the normal vector.

The cosine of the angle \( \theta' \) can be expressed using the dot product:
\[ \cos \theta' = \frac{\vec{l_1} \cdot \vec{n}}{|\vec{l_1}| |\vec{n}|} \]

Calculating the dot product:
\[ \vec{l_1} \cdot \vec{n} = (1)(2) + (2)(-1) + (2)(\sqrt{p}) = 2 - 2 + 2\sqrt{p} = 2\sqrt{p} \]

### Step 4: Calculate the magnitudes of the vectors.

The magnitude of \( \vec{l_1} \):
\[ |\vec{l_1}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

The magnitude of \( \vec{n} \):
\[ |\vec{n}| = \sqrt{2^2 + (-1)^2 + (\sqrt{p})^2} = \sqrt{4 + 1 + p} = \sqrt{5 + p} \]

### Step 5: Set up the equation using the cosine value.

Substituting into the cosine formula:
\[ \frac{2\sqrt{p}}{3\sqrt{5 + p}} = \frac{1}{3} \]

### Step 6: Solve for \( p \).

Cross-multiplying gives:
\[ 2\sqrt{p} = \sqrt{5 + p} \]
Squaring both sides:
\[ 4p = 5 + p \]
Rearranging:
\[ 4p - p = 5 \implies 3p = 5 \implies p = \frac{5}{3} \]

### Conclusion

The value of \( p \) is \( \frac{5}{3} \), which corresponds to option (D).

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