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

Answer

To solve the problem, we need to find the magnitude of the vector \(\vec{x}\) given the equation:

\[ (\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12 \]

where \(\vec{a}\) is a unit vector.

### Step-by-step Solution:

1. **Expand the Dot Product**:
We start by expanding the left-hand side of the equation:
\[ (\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = \vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{a} - \vec{a} \cdot \vec{x} - \vec{a} \cdot \vec{a} \]

2. **Simplify the Expression**:
Notice that \(\vec{x} \cdot \vec{a} = \vec{a} \cdot \vec{x}\). Thus, the terms \(\vec{x} \cdot \vec{a}\) and \(-\vec{a} \cdot \vec{x}\) cancel each other out:
\[ \vec{x} \cdot \vec{x} - \vec{a} \cdot \vec{a} = 12 \]

3. **Substitute the Magnitude of \(\vec{a}\)**:
Since \(\vec{a}\) is a unit vector, we know that:
\[ \vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1 \]
Therefore, substituting this into the equation gives:
\[ |\vec{x}|^2 - 1 = 12 \]

4. **Solve for \(|\vec{x}|^2\)**:
Rearranging the equation, we find:
\[ |\vec{x}|^2 = 12 + 1 = 13 \]

5. **Find \(|\vec{x}|\)**:
Finally, taking the square root to find the magnitude of \(\vec{x}\):
\[ |\vec{x}| = \sqrt{13} \]

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