We use the vector triple product expansion formula, which is \( \vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C} \).
Let's expand both sides of the given equation.
LHS: \( (\vec{a} \times \vec{b}) \times \vec{c} = -\vec{c} \times (\vec{a} \times \vec{b}) = -[(\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}] = (\vec{c} \cdot \vec{a})\vec{b} - (\vec{c} \cdot \vec{b})\vec{a} \).
RHS: \( \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} \).
Now, equate the expanded LHS and RHS:
\[ (\vec{c} \cdot \vec{a})\vec{b} - (\vec{c} \cdot \vec{b})\vec{a} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} \]
Since the dot product is commutative (\( \vec{a} \cdot \vec{c} = \vec{c} \cdot \vec{a} \)), the first term on both sides is identical and cancels out.
\[ -(\vec{c} \cdot \vec{b})\vec{a} = -(\vec{a} \cdot \vec{b})\vec{c} \]
\[ (\vec{b} \cdot \vec{c})\vec{a} = (\vec{a} \cdot \vec{b})\vec{c} \]
This equation shows that vector \( \vec{a} \) is a scalar multiple of vector \( \vec{c} \), where the scalar is \( k = \frac{\vec{a} \cdot \vec{b}}{\vec{b} \cdot \vec{c}} \).
If one vector is a scalar multiple of another (\( \vec{a} = k\vec{c} \)), it means the vectors are parallel or collinear.