The value of `[(veca-vecb, vecb-vecc, vecc-veca)]`, where `|veca|=1, |vecb|=5, |vecc|=3`, is

Options

A. `0`
B. `1`
C. `6`
D. none of these

Explanation

To solve the problem of finding the value of \([( \vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{c} - \vec{a})]\), where \(|\vec{a}| = 1\), \(|\vec{b}| = 5\), and \(|\vec{c}| = 3\), we will use the properties of the scalar triple product.

### Step-by-Step Solution:

1. **Understanding the Scalar Triple Product**:
The scalar triple product of three vectors \(\vec{u}, \vec{v}, \vec{w}\) is defined as:
\[ [\vec{u}, \vec{v}, \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w}) \]
In our case, we need to evaluate:
\[ [\vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{c} - \vec{a}] \]

2. **Expanding the Scalar Triple Product**:
We can express the scalar triple product as:
\[ [\vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{c} - \vec{a}] = (\vec{a} - \vec{b}) \cdot ((\vec{b} - \vec{c}) \times (\vec{c} - \vec{a})) \]

3. **Using Properties of Cross Product**:
We can expand the cross product:
\[ (\vec{b} - \vec{c}) \times (\vec{c} - \vec{a}) = \vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{c} \times \vec{c} + \vec{c} \times \vec{a} \]
Since \(\vec{c} \times \vec{c} = \vec{0}\), we have:
\[ (\vec{b} - \vec{c}) \times (\vec{c} - \vec{a}) = \vec{b} \times \vec{c} - \vec{b} \times \vec{a} + \vec{c} \times \vec{a} \]

4. **Substituting Back**:
Substitute this back into the scalar triple product:
\[ [\vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{c} - \vec{a}] = (\vec{a} - \vec{b}) \cdot (\vec{b} \times \vec{c} - \vec{b} \times \vec{a} + \vec{c} \times \vec{a}) \]

5. **Distributing the Dot Product**:
Now, distribute the dot product:
\[ = (\vec{a} - \vec{b}) \cdot (\vec{b} \times \vec{c}) - (\vec{a} - \vec{b}) \cdot (\vec{b} \times \vec{a}) + (\vec{a} - \vec{b}) \cdot (\vec{c} \times \vec{a}) \]

6. **Evaluating Each Term**:
- The term \((\vec{a} - \vec{b}) \cdot (\vec{b} \times \vec{c})\) remains.
- The term \((\vec{a} - \vec{b}) \cdot (\vec{b} \times \vec{a})\) is zero because \(\vec{b} \times \vec{a}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\).
- The term \((\vec{a} - \vec{b}) \cdot (\vec{c} \times \vec{a})\) is also zero for the same reason.

7. **Final Result**:
After simplifying, we find that all terms cancel out, leading to:
\[ [\vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{c} - \vec{a}] = 0 \]

### Conclusion:
Thus, the value of \([( \vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{c} - \vec{a})]\) is **0**.