Let $$\vec{a}, \vec{b}$$ and $$\vec{c}$$ be three non zero vectors such that $$\vec{b} \cdot \vec{c} = 0$$ and $$\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b} - \vec{c}}{2}$$. If $$\vec{d}$$ be a vector such that $$\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b}$$, then $$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d})$$ is equal to

We are given $$\vec{b} \cdot \vec{c} = 0$$ and $$\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b} - \vec{c}}{2}$$. Since the BAC-CAB identity states that $$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})$$, substituting yields the equation relating these expressions for the triple product.

Therefore, we have

$$\vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b}) = \frac{\vec{b} - \vec{c}}{2}\;. $$

Comparing coefficients of $$\vec{b}$$ and $$\vec{c}$$ from this equation gives $$\vec{a} \cdot \vec{c} = \frac{1}{2}$$ and $$\vec{a} \cdot \vec{b} = \frac{1}{2}$$. Next, we note that $$\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b} = \frac{1}{2}$$ and, from above, $$\vec{b} \cdot \vec{c} = 0$$.

Applying the scalar quadruple product identity,

$$ (\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})\;, $$

and substituting the known values leads to

$$ (\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) - (\vec{a} \cdot \vec{d})(0) = \frac{1}{4}\;. $$

Therefore, the correct answer is Option D: $$\frac{1}{4}$$.