Let ` hat(alpha) = 3 hat(i)+hat(j)` and `hat(beta)= 2 hat(i)-hat(j)+3hat(k)`. If `vec(beta)=vec(beta)_(1)-vec(beta)_(2)`, where `vec(beta)_(1)` is parallel to `vec(alpha)` and `vec(beta)_(2)` is perpendicular to `vec(alpha)` then `vec(beta)_(1)xx vec(beta)_(2)` is equal to :

Options

A. `- 3 hat(i) + 9 hat(j) + 5 hat(k)`
B. `1/2 ( 3 hat(i)- 9 hat(j) + 5 hat(k))`
C. `1/2 (-3 hat(i) + 9 hat(j) + 5 hat(k))`
D. `3 hat(i)-9 hat(j)- 5 hat(k)`

Explanation

To solve the problem, we need to find the cross product of two vectors, \( \vec{\beta_1} \) and \( \vec{\beta_2} \), where \( \vec{\beta_1} \) is parallel to \( \vec{\alpha} \) and \( \vec{\beta_2} \) is perpendicular to \( \vec{\alpha} \).

### Step-by-step Solution:

1. **Define the vectors**:
\[ \vec{\alpha} = 3\hat{i} + \hat{j} \]
\[ \vec{\beta} = 2\hat{i} - \hat{j} + 3\hat{k} \]

2. **Express \( \vec{\beta_1} \)**:
Since \( \vec{\beta_1} \) is parallel to \( \vec{\alpha} \), we can express it as:
\[ \vec{\beta_1} = \lambda \vec{\alpha} = \lambda (3\hat{i} + \hat{j}) = 3\lambda \hat{i} + \lambda \hat{j} \]

3. **Express \( \vec{\beta_2} \)**:
Given \( \vec{\beta} = \vec{\beta_1} - \vec{\beta_2} \), we can rearrange this to find \( \vec{\beta_2} \):
\[ \vec{\beta_2} = \vec{\beta_1} - \vec{\beta} = (3\lambda \hat{i} + \lambda \hat{j}) - (2\hat{i} - \hat{j} + 3\hat{k}) \]
Simplifying this gives:
\[ \vec{\beta_2} = (3\lambda - 2)\hat{i} + (\lambda + 1)\hat{j} - 3\hat{k} \]

4. **Condition for perpendicularity**:
Since \( \vec{\beta_2} \) is perpendicular to \( \vec{\alpha} \), we have:
\[ \vec{\alpha} \cdot \vec{\beta_2} = 0 \]
This leads to:
\[ (3\hat{i} + \hat{j}) \cdot ((3\lambda - 2)\hat{i} + (\lambda + 1)\hat{j} - 3\hat{k}) = 0 \]
Expanding this gives:
\[ 3(3\lambda - 2) + 1(\lambda + 1) + 0 = 0 \]
Simplifying:
\[ 9\lambda - 6 + \lambda + 1 = 0 \implies 10\lambda - 5 = 0 \implies \lambda = \frac{1}{2} \]

5. **Substituting back to find \( \vec{\beta_1} \) and \( \vec{\beta_2} \)**:
\[ \vec{\beta_1} = 3\left(\frac{1}{2}\right)\hat{i} + \left(\frac{1}{2}\right)\hat{j} = \frac{3}{2}\hat{i} + \frac{1}{2}\hat{j} \]
Now substitute \( \lambda \) into \( \vec{\beta_2} \):
\[ \vec{\beta_2} = \left(3\left(\frac{1}{2}\right) - 2\right)\hat{i} + \left(\frac{1}{2} + 1\right)\hat{j} - 3\hat{k} \]
Simplifying gives:
\[ \vec{\beta_2} = \left(\frac{3}{2} - 2\right)\hat{i} + \left(\frac{3}{2}\right)\hat{j} - 3\hat{k} = -\frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k} \]

6. **Calculate the cross product \( \vec{\beta_1} \times \vec{\beta_2} \)**:
Using the determinant form:
\[ \vec{\beta_1} \times \vec{\beta_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{3}{2} & \frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{3}{2} & -3 \end{vmatrix} \]
Expanding this determinant:
\[ = \hat{i} \left(\frac{1}{2} \cdot (-3) - 0 \cdot \frac{3}{2}\right) - \hat{j} \left(\frac{3}{2} \cdot (-3) - 0 \cdot -\frac{1}{2}\right) + \hat{k} \left(\frac{3}{2} \cdot \frac{3}{2} - (-\frac{1}{2}) \cdot \frac{1}{2}\right) \]
Simplifying gives:
\[ = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \left(\frac{9}{4} + \frac{1}{4}\right)\hat{k} = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \frac{10}{4}\hat{k} \]
Thus:
\[ = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \frac{5}{2}\hat{k} \]

### Final Answer:
\[ \vec{\beta_1} \times \vec{\beta_2} = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \frac{5}{2}\hat{k} \]