find `|vec x|` , if for a unit vector `vec a, (vec x- vec a)( vec x + vec a)`=12

Explanation

To solve the problem, we need to find the magnitude of the vector \(\vec{x}\) given the equation \((\vec{x} - \vec{a})(\vec{x} + \vec{a}) = 12\), where \(\vec{a}\) is a unit vector.

### Step-by-step Solution:

1. **Start with the given equation**:
\[ (\vec{x} - \vec{a})(\vec{x} + \vec{a}) = 12 \]

2. **Use the distributive property (FOIL method)**:
\[ \vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{a} - \vec{a} \cdot \vec{x} - \vec{a} \cdot \vec{a} = 12 \]

3. **Simplify the equation**:
- The terms \(\vec{x} \cdot \vec{a}\) and \(-\vec{a} \cdot \vec{x}\) cancel each other out.
- The term \(\vec{a} \cdot \vec{a}\) is equal to \(|\vec{a}|^2\). Since \(\vec{a}\) is a unit vector, \(|\vec{a}|^2 = 1\).
- Thus, we have:
\[ \vec{x} \cdot \vec{x} - 1 = 12 \]

4. **Rearrange the equation**:
\[ \vec{x} \cdot \vec{x} = 12 + 1 \]
\[ \vec{x} \cdot \vec{x} = 13 \]

5. **Find the magnitude of \(\vec{x}\)**:
- The magnitude of a vector \(\vec{x}\) is given by \(|\vec{x}| = \sqrt{\vec{x} \cdot \vec{x}}\).
- Therefore:
\[ |\vec{x}| = \sqrt{13} \]

### Final Answer:
\[ |\vec{x}| = \sqrt{13} \]