If the acute angle that the vector `alphahati+betahatj+gammahatk` makes with the plane of the two vectors `2hati+3hatj-hatk` and `hati-hatj+2hatk` is `tan^(-1)(1/(sqrt(2)))` then

Options

A. `alpha(beta+gamma)=beta gamma`
B. `beta(gamma+alpha)=gamma alpha`
C. `gamma(alpha+beta)=alpha beta`
D. `alpha beta =beta gamma +gamma alpha =0`

Explanation

To solve the problem, we need to find the relationship between the given vectors and the angle they make with the plane formed by the two vectors. We are given the vector \( \mathbf{R} = \hat{i} + \hat{j} + \hat{k} \) and the two vectors \( \mathbf{A} = 2\hat{i} + 3\hat{j} - \hat{k} \) and \( \mathbf{B} = \hat{i} - \hat{j} + 2\hat{k} \). The acute angle \( \theta \) that \( \mathbf{R} \) makes with the plane formed by \( \mathbf{A} \) and \( \mathbf{B} \) is given as \( \tan^{-1}\left(\frac{1}{\sqrt{2}}\right) \).

### Step-by-Step Solution:

1. **Find the Cross Product of Vectors A and B**:
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & -1 & 2 \end{vmatrix} \]
Expanding this determinant:
\[ \mathbf{A} \times \mathbf{B} = \hat{i} \left(3 \cdot 2 - (-1) \cdot (-1)\right) - \hat{j} \left(2 \cdot 2 - (-1) \cdot 1\right) + \hat{k} \left(2 \cdot (-1) - 3 \cdot 1\right) \]
\[ = \hat{i}(6 - 1) - \hat{j}(4 + 1) + \hat{k}(-2 - 3) \]
\[ = 5\hat{i} - 5\hat{j} - 5\hat{k} \]
\[ = 5(\hat{i} - \hat{j} - \hat{k}) \]

2. **Find the Magnitude of the Cross Product**:
\[ |\mathbf{A} \times \mathbf{B}| = 5 \sqrt{1^2 + (-1)^2 + (-1)^2} = 5 \sqrt{3} \]

3. **Find the Magnitude of Vector R**:
\[ |\mathbf{R}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]

4. **Use the Given Angle to Find the Relationship**:
The angle \( \theta \) that \( \mathbf{R} \) makes with the normal to the plane is given by:
\[ \sin(\theta) = \frac{|\mathbf{R} \cdot (\mathbf{A} \times \mathbf{B})|}{|\mathbf{R}| |\mathbf{A} \times \mathbf{B}|} \]
Given \( \theta = \tan^{-1}\left(\frac{1}{\sqrt{2}}\right) \), we find \( \sin(\theta) \):
\[ \sin(\theta) = \frac{1}{\sqrt{3}} \]

5. **Set Up the Equation**:
\[ \frac{|\mathbf{R} \cdot (\mathbf{A} \times \mathbf{B})|}{\sqrt{3} \cdot 5\sqrt{3}} = \frac{1}{\sqrt{3}} \]
Simplifying gives:
\[ |\mathbf{R} \cdot (\mathbf{A} \times \mathbf{B})| = 5 \]

6. **Calculate the Dot Product**:
\[ \mathbf{R} \cdot (\mathbf{A} \times \mathbf{B}) = (1)(5) + (1)(-5) + (1)(-5) = 5 - 5 - 5 = -5 \]
Therefore, \( |\mathbf{R} \cdot (\mathbf{A} \times \mathbf{B})| = 5 \).

7. **Conclusion**:
Since the absolute value of the dot product matches our earlier calculation, we confirm that the acute angle condition holds true.