`veca ,vecb and vecc` are unimodular and coplanar. A unit vector `vecd` is perpendicualt to them , `(veca xx vecb) xx (vecc xx vecd) = 1/6 hati - 1/3 hatj + 1/3 hatk` , and the angle between `veca and vecb is 30^(@)` then `vecc` is

Options

A. `(hati-2 hatj + 2 hatk)//3`
B. `(-hati +2hatj - 2 hatk)//3`
C. `(-hati +2hatj - hatk)//3`
D. `(-2hati -2hatj + hatk)//3`

Explanation

To solve the problem step by step, we will use the properties of vectors, particularly focusing on the scalar triple product and the geometric relationships given in the problem.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Vectors \( \vec{a}, \vec{b}, \vec{c} \) are unimodular (each has a magnitude of 1).
- Vectors \( \vec{a}, \vec{b}, \vec{c} \) are coplanar, which implies that their scalar triple product is zero: \( \vec{a} \cdot (\vec{b} \times \vec{c}) = 0 \).
- A unit vector \( \vec{d} \) is perpendicular to \( \vec{a}, \vec{b}, \) and \( \vec{c} \).
- The equation given is \( (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \frac{1}{6} \hat{i} - \frac{1}{3} \hat{j} + \frac{1}{3} \hat{k} \).
- The angle between \( \vec{a} \) and \( \vec{b} \) is \( 30^\circ \).

2. **Using the Scalar Triple Product**:
- The scalar triple product can be expressed as \( \vec{a} \cdot (\vec{b} \times \vec{c}) \).
- Since \( \vec{d} \) is perpendicular to \( \vec{a} \) and \( \vec{b} \), we can express \( \vec{c} \) in terms of \( \vec{a} \) and \( \vec{b} \).

3. **Calculating the Cross Product**:
- The expression \( (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) \) can be simplified using the vector triple product identity:
\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \]
- Here, let \( \vec{x} = \vec{a} \times \vec{b} \), \( \vec{y} = \vec{c} \), and \( \vec{z} = \vec{d} \).

4. **Finding the Magnitude**:
- The magnitude of \( \vec{a} \times \vec{b} \) is given by:
\[ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin(30^\circ) = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2} \]
- Thus, \( \vec{a} \times \vec{b} \) is a vector with magnitude \( \frac{1}{2} \).

5. **Substituting into the Equation**:
- We can substitute into the equation:
\[ \frac{1}{2} \cdot \vec{c} \cdot \vec{d} = \frac{1}{6} \hat{i} - \frac{1}{3} \hat{j} + \frac{1}{3} \hat{k} \]
- Since \( \vec{d} \) is a unit vector, we can equate:
\[ \vec{c} \cdot \vec{d} = \frac{1}{6} \cdot 2 \hat{i} - \frac{1}{3} \cdot 2 \hat{j} + \frac{1}{3} \cdot 2 \hat{k} = \frac{1}{3} \hat{i} - \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k} \]

6. **Finding \( \vec{c} \)**:
- Since \( \vec{c} \) is also a unit vector, we can express \( \vec{c} \) in terms of its components:
\[ \vec{c} = x \hat{i} + y \hat{j} + z \hat{k} \]
- Using the fact that \( |\vec{c}| = 1 \), we have:
\[ x^2 + y^2 + z^2 = 1 \]

7. **Solving for \( \vec{c} \)**:
- From the previous calculations, we can find \( \vec{c} \) that satisfies both the dot product and the unit vector condition.
- After solving, we find:
\[ \vec{c} = \frac{1}{3} \hat{i} - \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k} \]

### Final Answer:
\[ \vec{c} = \frac{1}{3} \hat{i} - \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k} \]