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. (a) 0
B. (b) 1
C. (c) `1/4(a_(1)^(2)+a_(2)^(2)+a_(2)^(2))(b_(1)^(2) +b_(2)^(2)+b_(2)^(2))`
D. (d)`3/4(a_(1)^(2)+a_(2)^(2)+a_(2)^(2))(b_(1)^(2) +b_(2)^(2)+b_(2)^(2)) (c_(1)^(2) + c_(2)^(2)+c_(2)^(2))`

Explanation

To solve the problem, we need to find the square of the scalar triple product of the vectors \( \overset{\to}{a} \), \( \overset{\to}{b} \), and \( \overset{\to}{c} \). Here are the steps to arrive at the solution:

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

### Step 2: Use the property of the scalar triple product
The scalar triple product can be expressed as:
\[ |(\overset{\to}{a}, \overset{\to}{b}, \overset{\to}{c})| = \overset{\to}{a} \cdot (\overset{\to}{b} \times \overset{\to}{c}) \]
This can also be written as:
\[ |(\overset{\to}{a}, \overset{\to}{b}, \overset{\to}{c})|^2 = |\overset{\to}{a}|^2 |\overset{\to}{b}|^2 |\overset{\to}{c}|^2 \sin^2 \theta \]
where \( \theta \) is the angle between vectors \( \overset{\to}{a} \) and \( \overset{\to}{b} \).

### Step 3: Calculate the sine of the angle
Given that the angle \( \theta \) between \( \overset{\to}{a} \) and \( \overset{\to}{b} \) is \( \frac{\pi}{6} \):
\[ \sin \theta = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} \]

### Step 4: Substitute the values into the formula
Since \( \overset{\to}{c} \) is a unit vector, we have:
\[ |\overset{\to}{c}| = 1 \implies |\overset{\to}{c}|^2 = 1 \]
Thus, the expression simplifies to:
\[ |(\overset{\to}{a}, \overset{\to}{b}, \overset{\to}{c})|^2 = |\overset{\to}{a}|^2 |\overset{\to}{b}|^2 \sin^2 \left(\frac{\pi}{6}\right) \]
\[ = |\overset{\to}{a}|^2 |\overset{\to}{b}|^2 \left(\frac{1}{2}\right)^2 = |\overset{\to}{a}|^2 |\overset{\to}{b}|^2 \cdot \frac{1}{4} \]

### Step 5: Express the magnitudes of vectors
Let:
\[ |\overset{\to}{a}|^2 = a_1^2 + a_2^2 + a_3^2 \]
\[ |\overset{\to}{b}|^2 = b_1^2 + b_2^2 + b_3^2 \]
Then:
\[ |(\overset{\to}{a}, \overset{\to}{b}, \overset{\to}{c})|^2 = \frac{1}{4} (a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2) \]

### Conclusion
Thus, the final result for \( |(\overset{\to}{a}, \overset{\to}{b}, \overset{\to}{c})|^2 \) is:
\[ |(\overset{\to}{a}, \overset{\to}{b}, \overset{\to}{c})|^2 = \frac{1}{4} \cdot |(a_1, a_2, a_3), (b_1, b_2, b_3)|^2 \]