Let three vectors $$\vec{a}, \vec{b}$$ and $$\vec{c}$$ be such that $$\vec{a} \times \vec{b} = \vec{c}$$, $$\vec{b} \times \vec{c} = \vec{a}$$ and $$|\vec{a}| = 2$$. Then which one of the following is not true?

We are given $$\vec{a} \times \vec{b} = \vec{c}$$, $$\vec{b} \times \vec{c} = \vec{a}$$, and $$|\vec{a}| = 2$$. First, let us establish the magnitudes and mutual orthogonality of the three vectors.

From $$\vec{b} \times \vec{c} = \vec{a}$$, take the cross product with $$\vec{b}$$ from the left: $$\vec{b} \times (\vec{b} \times \vec{c}) = \vec{b} \times \vec{a}$$. Using the BAC-CAB identity on the left: $$(\vec{b} \cdot \vec{c})\vec{b} - |\vec{b}|^2 \vec{c} = -\vec{a} \times \vec{b} = -\vec{c}$$. This gives $$(\vec{b} \cdot \vec{c})\vec{b} = (|\vec{b}|^2 - 1)\vec{c}$$. Since $$\vec{b}$$ and $$\vec{c}$$ are linearly independent (their cross product $$\vec{a} \neq \vec{0}$$), both coefficients must be zero: $$\vec{b} \cdot \vec{c} = 0$$ and $$|\vec{b}|^2 = 1$$, giving $$|\vec{b}| = 1$$.

Since $$\vec{b} \cdot \vec{c} = 0$$, from $$\vec{b} \times \vec{c} = \vec{a}$$ we get $$|\vec{a}| = |\vec{b}||\vec{c}|\sin 90° = |\vec{c}|$$, so $$|\vec{c}| = 2$$. From $$\vec{a} \times \vec{b} = \vec{c}$$, $$|\vec{c}| = |\vec{a}||\vec{b}|\sin\theta_{ab} = 2\sin\theta_{ab} = 2$$, giving $$\sin\theta_{ab} = 1$$, so $$\vec{a} \perp \vec{b}$$. Also, $$\vec{c} = \vec{a} \times \vec{b}$$ is perpendicular to both $$\vec{a}$$ and $$\vec{b}$$. Therefore $$\vec{a}, \vec{b}, \vec{c}$$ are mutually orthogonal with $$|\vec{a}| = |\vec{c}| = 2$$ and $$|\vec{b}| = 1$$.

Now check each option. For Option (A): $$(\vec{b} + \vec{c}) \times (\vec{b} - \vec{c}) = \vec{b} \times \vec{b} - \vec{b} \times \vec{c} + \vec{c} \times \vec{b} - \vec{c} \times \vec{c} = -2(\vec{b} \times \vec{c}) = -2\vec{a}$$. So $$\vec{a} \times (-2\vec{a}) = \vec{0}$$. This is true.

For Option (B): the projection of $$\vec{a}$$ on $$\vec{b} \times \vec{c}$$ is $$\frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|}$$. Since $$\vec{b} \times \vec{c} = \vec{a}$$, this is $$\frac{|\vec{a}|^2}{|\vec{a}|} = |\vec{a}| = 2$$. This is true.

For Option (C): $$[\vec{a}\;\vec{b}\;\vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{a} \cdot \vec{a} = 4$$. The scalar triple product is invariant under cyclic permutation, so $$[\vec{c}\;\vec{a}\;\vec{b}] = 4$$ as well. Their sum is $$4 + 4 = 8$$. This is true.

For Option (D): $$|3\vec{a} + \vec{b} - 2\vec{c}|^2 = 9|\vec{a}|^2 + |\vec{b}|^2 + 4|\vec{c}|^2 + 6(\vec{a} \cdot \vec{b}) - 12(\vec{a} \cdot \vec{c}) - 4(\vec{b} \cdot \vec{c})$$. Since all dot products are zero, this equals $$9(4) + 1 + 4(4) = 36 + 1 + 16 = 53$$. The option claims the value is 51, which is incorrect.

Therefore Option (D) is not true, and the answer is $$|3\vec{a} + \vec{b} - 2\vec{c}|^2 = 53 \neq 51$$.