Let ` overset(to)(a) =a_(1) hat(i) + a_(2) hat(j) + a_(3) hat(k) , overset(to)(b) = b_(1) hat(i) +b_(2) hat(j) +b_(3) hat(k) " and " overset(to)(c) = c_(1) hat(i) +c_(2) hat(j) + c_(3) hat(k)` be three non- zero vectors such that `overset(to)(c )` is a unit vectors perpendicular to both the vectors `overset(to)(a )` and `overset(to)(b)`. If the angle between `overset(to)(a) " and " overset(to)(b)` 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) (a_(1)^(2) +a_(2)^(2) +a_(3)^(2))(b_(1)^(2)+b_(1)^(2) +b_(3)^(2))`
D. `(3)/(4) (a_(1)^(2)+a_(2)^(2))(b_(1)^(2)+b_(2)^(2)+a_(3)^(2))(c_(1)^(2)+c_(2)^(2)+c_(3)^(2))`

Explanation

To solve the given problem, we need to find the value of the determinant \( |(a_1, a_2, a_3), (b_1, b_2, b_3), (c_1, c_2, c_3)|^2 \) given the conditions about the vectors \( \overset{\to}{a} \), \( \overset{\to}{b} \), and \( \overset{\to}{c} \).

### Step-by-Step Solution:

1. **Understanding the Vectors**:
- Let \( \overset{\to}{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \)
- Let \( \overset{\to}{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \)
- Let \( \overset{\to}{c} = c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k} \)

2. **Given Conditions**:
- \( \overset{\to}{c} \) is a unit vector perpendicular to both \( \overset{\to}{a} \) and \( \overset{\to}{b} \).
- The angle \( \theta \) between \( \overset{\to}{a} \) and \( \overset{\to}{b} \) is \( \frac{\pi}{6} \).

3. **Cross Product**:
- The cross product \( \overset{\to}{a} \times \overset{\to}{b} \) gives a vector that is perpendicular to both \( \overset{\to}{a} \) and \( \overset{\to}{b} \).
- The magnitude of the cross product is given by:
\[ |\overset{\to}{a} \times \overset{\to}{b}| = |\overset{\to}{a}| |\overset{\to}{b}| \sin(\theta) \]
- Since \( \theta = \frac{\pi}{6} \), we have \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \).

4. **Magnitude of the Cross Product**:
- Let \( |\overset{\to}{a}| = A \) and \( |\overset{\to}{b}| = B \).
- Thus, the magnitude becomes:
\[ |\overset{\to}{a} \times \overset{\to}{b}| = A \cdot B \cdot \frac{1}{2} \]

5. **Using the Unit Vector**:
- Since \( \overset{\to}{c} \) is a unit vector, we can express it as:
\[ \overset{\to}{c} = \frac{\overset{\to}{a} \times \overset{\to}{b}}{|\overset{\to}{a} \times \overset{\to}{b}|} \]

6. **Triple Product**:
- The scalar triple product \( \overset{\to}{a} \cdot (\overset{\to}{b} \times \overset{\to}{c}) \) gives the volume of the parallelepiped formed by the vectors.
- This can also be expressed as:
\[ |(\overset{\to}{a}, \overset{\to}{b}, \overset{\to}{c})| = |\overset{\to}{a}| |\overset{\to}{b}| |\overset{\to}{c}| \cdot \cos(\phi) \]
- Where \( \phi \) is the angle between \( \overset{\to}{c} \) and \( \overset{\to}{a} \).

7. **Calculating the Determinant**:
- Since \( \overset{\to}{c} \) is perpendicular to both \( \overset{\to}{a} \) and \( \overset{\to}{b} \), the angle \( \phi = 90^\circ \) implies \( \cos(90^\circ) = 0 \).
- Therefore:
\[ |(\overset{\to}{a}, \overset{\to}{b}, \overset{\to}{c})| = \frac{1}{2} A B \]

8. **Final Calculation**:
- The square of the determinant is:
\[ |(\overset{\to}{a}, \overset{\to}{b}, \overset{\to}{c})|^2 = \left(\frac{1}{2} A B\right)^2 = \frac{1}{4} A^2 B^2 \]

### Conclusion:
The final result is:
\[ |(\overset{\to}{a}, \overset{\to}{b}, \overset{\to}{c})|^2 = \frac{1}{4} |A|^2 |B|^2 \]