If $$\vec{a} \cdot \vec{b} = 1, \vec{b} \cdot \vec{c} = 2$$ and $$\vec{c} \cdot \vec{a} = 3$$, then the value of $$\left[\vec{a} \times (\vec{b} \times \vec{c}), \vec{b} \times (\vec{c} \times \vec{a}), \vec{c} \times (\vec{b} \times \vec{a})\right]$$ is

The given information is $$\vec{a} \cdot \vec{b} = 1$$, $$\vec{b} \cdot \vec{c} = 2$$, and $$\vec{c} \cdot \vec{a} = 3$$, and we seek $$[\vec{a} \times (\vec{b} \times \vec{c}),\ \vec{b} \times (\vec{c} \times \vec{a}),\ \vec{c} \times (\vec{b} \times \vec{a})].$$

Using the BAC-CAB identity, we obtain $$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b}) = 3\vec{b} - \vec{c},$$ $$\vec{b} \times (\vec{c} \times \vec{a}) = \vec{c}(\vec{b} \cdot \vec{a}) - \vec{a}(\vec{b} \cdot \vec{c}) = \vec{c} - 2\vec{a},$$ and $$\vec{c} \times (\vec{b} \times \vec{a}) = \vec{b}(\vec{c} \cdot \vec{a}) - \vec{a}(\vec{c} \cdot \vec{b}) = 3\vec{b} - 2\vec{a}.$$

Letting $$\vec{u} = 3\vec{b} - \vec{c},$$ $$\vec{v} = \vec{c} - 2\vec{a},$$ and $$\vec{w} = 3\vec{b} - 2\vec{a},$$ the scalar triple product is defined by $$[\vec{u}, \vec{v}, \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w}).$$

Observing that $$\vec{u} + \vec{v} = (3\vec{b} - \vec{c}) + (\vec{c} - 2\vec{a}) = 3\vec{b} - 2\vec{a} = \vec{w},$$ the property of the scalar triple product gives $$[\vec{u}, \vec{v}, \vec{u} + \vec{v}] = [\vec{u}, \vec{v}, \vec{u}] + [\vec{u}, \vec{v}, \vec{v}],$$ both of which vanish since the scalar triple product is zero when two vectors coincide.

Therefore, $$[\vec{u}, \vec{v}, \vec{w}] = 0$$ and the correct answer is Option A: $$0$$.