`veca, vecb and vecc` are coplanar unit vectors. A unit vector `vecd` is perpendicular to them. If `(vecaxxvecb)xx(vecc xx vecb)=(3)/(26)hati-(2)/(13)hatj+(6)/(13)hatk`
and the angle between `veca and vecb` is `30^(@)`, then `vecc` is equal to

Options

A. `(3)/(13)hati-(4)/(13)hatj+(12)/(13)hatk`
B. `(2)/(7)hati-(3)/(7)hatj+(6)/(7)hatk`
C. `3hati-4hatj+12hatk`
D. `(1)/(sqrt3)hati-(1)/(sqrt3)hatj+(1)/(sqrt3)hatk`

Answer

To solve the problem step by step, we need to find the vector \(\vec{c}\) given the conditions of the problem.

### Step 1: Understand the Given Information
We know that \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are coplanar unit vectors, and \(\vec{d}\) is a unit vector that is perpendicular to all three. We are given the expression:
\[ (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{b}) = \frac{3}{26} \hat{i} - \frac{2}{13} \hat{j} + \frac{6}{13} \hat{k} \]
and the angle between \(\vec{a}\) and \(\vec{b}\) is \(30^\circ\).

### Step 2: Use the Cross Product Identity
Using the vector triple product identity:
\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \]
we can rewrite:
\[ (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{b}) = (\vec{a} \cdot \vec{b}) \vec{c} - (\vec{a} \cdot \vec{c}) \vec{b} \]

### Step 3: Calculate \(\vec{a} \cdot \vec{b}\)
Since the angle between \(\vec{a}\) and \(\vec{b}\) is \(30^\circ\), we can calculate:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(30^\circ) = 1 \cdot 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \]

### Step 4: Substitute into the Equation
Substituting the dot product into the equation gives:
\[ \frac{\sqrt{3}}{2} \vec{c} - (\vec{a} \cdot \vec{c}) \vec{b} = \frac{3}{26} \hat{i} - \frac{2}{13} \hat{j} + \frac{6}{13} \hat{k} \]

### Step 5: Isolate \(\vec{c}\)
Rearranging gives:
\[ \frac{\sqrt{3}}{2} \vec{c} = \frac{3}{26} \hat{i} - \frac{2}{13} \hat{j} + \frac{6}{13} \hat{k} + (\vec{a} \cdot \vec{c}) \vec{b} \]

### Step 6: Normalize the Expression
Since \(\vec{c}\) is a unit vector, we can express it as:
\[ \vec{c} = k \left( \frac{3}{26} \hat{i} - \frac{2}{13} \hat{j} + \frac{6}{13} \hat{k} \right) \]
where \(k\) is a scalar that we need to determine.

### Step 7: Find Magnitude
The magnitude of the right-hand side must equal 1 (since \(\vec{c}\) is a unit vector):
\[ k \sqrt{\left( \frac{3}{26} \right)^2 + \left( -\frac{2}{13} \right)^2 + \left( \frac{6}{13} \right)^2} = 1 \]

Calculating the magnitude:
\[ \sqrt{\left( \frac{3}{26} \right)^2 + \left( -\frac{2}{13} \right)^2 + \left( \frac{6}{13} \right)^2} = \sqrt{\frac{9}{676} + \frac{4}{169} + \frac{36}{169}} = \sqrt{\frac{9}{676} + \frac{40}{169}} = \sqrt{\frac{9 + 160}{676}} = \sqrt{\frac{169}{676}} = \frac{13}{26} = \frac{1}{2} \]

### Step 8: Solve for \(k\)
Thus:
\[ k \cdot \frac{1}{2} = 1 \implies k = 2 \]

### Step 9: Find \(\vec{c}\)
Substituting \(k\) back gives:
\[ \vec{c} = 2 \left( \frac{3}{26} \hat{i} - \frac{2}{13} \hat{j} + \frac{6}{13} \hat{k} \right) = \frac{6}{26} \hat{i} - \frac{4}{13} \hat{j} + \frac{12}{13} \hat{k} \]
Simplifying this gives:
\[ \vec{c} = \frac{3}{13} \hat{i} - \frac{4}{13} \hat{j} + \frac{12}{13} \hat{k} \]

### Final Answer
Thus, the vector \(\vec{c}\) is:
\[ \vec{c} = \frac{3}{13} \hat{i} - \frac{4}{13} \hat{j} + \frac{12}{13} \hat{k} \]