Let `veca=a_(1)hati+a_(2)hatj+a_(3)hatk, vecb=b_(1)hati+b_(2)hatj+b_(3)hatk` and `vecc=c_(1)hati+c_(2)hatj+c_(3)hatk` be three non zero vectors such that `vecc` is a unit vector perpendicular to both `veca` and `vecb` . If the angle between `veca` and `vecb` is `(pi)/6`, then `|(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3))|^(2)` is equal to

Options

A. `0`
B. `1`
C. `1/4|veca|^(2)|vecb|^(2)`
D. `3/4|veca|^(2)|vecb|^(2)`

Explanation

To solve the problem, we need to find the square of the scalar triple product of the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). Given the conditions in the problem, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Given Vectors**:
We have three vectors:
\[ \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \]
\[ \vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \]
\[ \vec{c} = c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k} \]
where \(\vec{c}\) is a unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\).

2. **Using the Property of the Scalar Triple Product**:
The scalar triple product can be expressed as:
\[ |\vec{a} \cdot (\vec{b} \times \vec{c})| = |\vec{a}| |\vec{b}| |\vec{c}| \sin(\theta) \]
where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\). Given that \(\theta = \frac{\pi}{6}\), we can express \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\).

3. **Magnitude of the Cross Product**:
The magnitude of the cross product \(|\vec{b} \times \vec{c}|\) can be expressed as:
\[ |\vec{b} \times \vec{c}| = |\vec{b}| |\vec{c}| \sin(\phi) \]
where \(\phi\) is the angle between \(\vec{b}\) and \(\vec{c}\). Since \(\vec{c}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\), \(\phi = \frac{\pi}{2}\) and \(\sin\left(\frac{\pi}{2}\right) = 1\).

4. **Calculating the Scalar Triple Product**:
The scalar triple product can also be expressed as:
\[ |\vec{a} \cdot (\vec{b} \times \vec{c})| = |\vec{a}| |\vec{b}| |\vec{c}| \sin\left(\frac{\pi}{6}\right) \]
Since \(|\vec{c}| = 1\) (as it is a unit vector), we have:
\[ |\vec{a} \cdot (\vec{b} \times \vec{c})| = |\vec{a}| |\vec{b}| \cdot \frac{1}{2} \]

5. **Finding the Square of the Scalar Triple Product**:
The square of the scalar triple product is:
\[ |\vec{a} \cdot (\vec{b} \times \vec{c})|^2 = \left( |\vec{a}| |\vec{b}| \cdot \frac{1}{2} \right)^2 = \frac{1}{4} |\vec{a}|^2 |\vec{b}|^2 \]

6. **Final Expression**:
Thus, the final expression for the square of the scalar triple product is:
\[ |\vec{a}, \vec{b}, \vec{c}|^2 = \frac{1}{4} |\vec{a}|^2 |\vec{b}|^2 \]

### Conclusion:
The value of \(|(\vec{a}, \vec{b}, \vec{c})|^2\) is equal to \(\frac{1}{4} |\vec{a}|^2 |\vec{b}|^2\).