To solve the problem, we need to show that if a non-zero vector \(\vec{x}\) is orthogonal to three vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), then the scalar triple product \([\vec{a}, \vec{b}, \vec{c}] = 0\).
### Step-by-Step Solution:
1. **Understanding the Given Conditions**:
We have the following conditions:
\[
\vec{x} \cdot \vec{a} = 0, \quad \vec{x} \cdot \vec{b} = 0, \quad \vec{x} \cdot \vec{c} = 0
\]
This means that the vector \(\vec{x}\) is orthogonal to the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).
2. **Analyzing the Implications**:
Since \(\vec{x}\) is orthogonal to \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), it implies that all three vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) lie in the same plane. This is because if a vector is orthogonal to multiple vectors, those vectors must be coplanar.
3. **Using the Definition of Scalar Triple Product**:
The scalar triple product \([\vec{a}, \vec{b}, \vec{c}]\) is defined as the volume of the parallelepiped formed by the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). If the vectors are coplanar, the volume is zero:
\[
[\vec{a}, \vec{b}, \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c})
\]
4. **Conclusion**:
Since \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are coplanar, the scalar triple product must equal zero:
\[
[\vec{a}, \vec{b}, \vec{c}] = 0
\]
Thus, we have shown that \([\vec{a}, \vec{b}, \vec{c}] = 0\).
