Unit vectors `vec(a),vec(b),vec(c )` are coplanar. A unit vector `vec(d)` is perpendicular to them. If `(vec(a)xxvec(b))xx(vec(c )xxvec(d))=(1)/(6)i-(1)/(3)hat(j)+(1)/(3)hat(k)` and the angle between `vec(a)` and `vec(b)` is `30^(@)`, then `vec(c )` is/are :

Options

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

Explanation

To solve the problem, we need to find the unit vector \(\vec{c}\) given that the unit vectors \(\vec{a}, \vec{b}, \vec{c}\) are coplanar and a unit vector \(\vec{d}\) is perpendicular to them. We also have the equation:

\[ (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \frac{1}{6} \hat{i} - \frac{1}{3} \hat{j} + \frac{1}{3} \hat{k} \]

and the angle between \(\vec{a}\) and \(\vec{b}\) is \(30^\circ\).

### Step 1: Understand the Coplanarity and Perpendicularity
Since \(\vec{a}, \vec{b}, \vec{c}\) are coplanar, we can express the cross product \(\vec{a} \times \vec{b}\) as a vector that is perpendicular to the plane formed by \(\vec{a}\) and \(\vec{b}\). The vector \(\vec{d}\) is also perpendicular to this plane.

### Step 2: Calculate \(\vec{a} \times \vec{b}\)
Using the formula for the magnitude of the cross product:

\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(\theta) \]

Given that \(\vec{a}\) and \(\vec{b}\) are unit vectors and \(\theta = 30^\circ\):

\[ |\vec{a} \times \vec{b}| = 1 \cdot 1 \cdot \sin(30^\circ) = \frac{1}{2} \]

Thus, we can write:

\[ \vec{a} \times \vec{b} = \frac{1}{2} \hat{n} \]

where \(\hat{n}\) is a unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\).

### Step 3: Use the Vector Triple Product Identity
We can apply the vector triple product identity:

\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \]

Let \(\vec{x} = \vec{a} \times \vec{b}\), \(\vec{y} = \vec{c}\), and \(\vec{z} = \vec{d}\):

\[ (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = (\vec{a} \times \vec{b}) \cdot \vec{d} \cdot \vec{c} - (\vec{a} \times \vec{b}) \cdot \vec{c} \cdot \vec{d} \]

Since \(\vec{d}\) is perpendicular to \(\vec{a}\) and \(\vec{b}\), we have:

\[ (\vec{a} \times \vec{b}) \cdot \vec{d} = 0 \]

Thus, the equation simplifies to:

\[ - (\vec{a} \times \vec{b}) \cdot \vec{c} \cdot \vec{d} = \frac{1}{6} \hat{i} - \frac{1}{3} \hat{j} + \frac{1}{3} \hat{k} \]

### Step 4: Solve for \(\vec{c}\)
Now we need to find \(\vec{c}\). We know:

\[ (\vec{a} \times \vec{b}) \cdot \vec{c} = \frac{1}{2} \hat{n} \cdot \vec{c} \]

Let \(\vec{c} = x \hat{i} + y \hat{j} + z \hat{k}\). We can express the dot product in terms of the components of \(\vec{c}\).

### Step 5: Substitute and Solve
We substitute and solve the equation:

\[ \frac{1}{2} (x \hat{i} + y \hat{j} + z \hat{k}) = \frac{1}{6} \hat{i} - \frac{1}{3} \hat{j} + \frac{1}{3} \hat{k} \]

This leads to a system of equations for \(x, y, z\):

1. \(\frac{1}{2} x = \frac{1}{6} \Rightarrow x = \frac{1}{3}\)
2. \(\frac{1}{2} y = -\frac{1}{3} \Rightarrow y = -\frac{2}{3}\)
3. \(\frac{1}{2} z = \frac{1}{3} \Rightarrow z = \frac{2}{3}\)

Thus, we have:

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

### Final Answer
The unit vector \(\vec{c}\) is:

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