Let $$\overrightarrow{a}=-\widehat{i}+\widehat{j}+2\widehat{k},\overrightarrow{b}=\widehat{i}-\widehat{j}-3\widehat{k},\overrightarrow{c}=\overrightarrow{a} \times \overrightarrow{b}\text{ and }\overrightarrow{d}=\overrightarrow{c}\times\overrightarrow{a}$$. Then $$\large (\overrightarrow{a}-\overrightarrow{b}).\overrightarrow{d}$$ is equal to:

We are given $$\vec{a} = -\hat{i} + \hat{j} + 2\hat{k}$$, $$\vec{b} = \hat{i} - \hat{j} - 3\hat{k}$$, $$\vec{c} = \vec{a} \times \vec{b}$$, and $$\vec{d} = \vec{c} \times \vec{a}$$. We need to find $$(\vec{a} - \vec{b}) \cdot \vec{d}$$.

First we compute $$\vec{c} = \vec{a} \times \vec{b}$$. Using the determinant form, we have $$\vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 2 \\ 1 & -1 & -3 \end{vmatrix}$$ $$= \hat{i}(1 \cdot (-3) - 2 \cdot (-1)) - \hat{j}((-1)(-3) - 2 \cdot 1) + \hat{k}((-1)(-1) - 1 \cdot 1)$$ $$= \hat{i}(-3 + 2) - \hat{j}(3 - 2) + \hat{k}(1 - 1) = -\hat{i} - \hat{j} + 0\hat{k}$$ so $$\vec{c} = (-1, -1, 0)$$.

Next, we compute $$\vec{d} = \vec{c} \times \vec{a}$$. Again using the determinant, $$\vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 0 \\ -1 & 1 & 2 \end{vmatrix}$$ $$= \hat{i}((-1)(2) - 0 \cdot 1) - \hat{j}((-1)(2) - 0 \cdot (-1)) + \hat{k}((-1)(1) - (-1)(-1))$$ $$= \hat{i}(-2) - \hat{j}(-2) + \hat{k}(-1 - 1) = -2\hat{i} + 2\hat{j} - 2\hat{k}$$ which gives $$\vec{d} = (-2, 2, -2)$$.

Then $$\vec{a} - \vec{b} = (-1 - 1, 1 - (-1), 2 - (-3)) = (-2, 2, 5)\,, $$ and the dot product is $$(\vec{a} - \vec{b}) \cdot \vec{d} = (-2)(-2) + (2)(2) + (5)(-2) = 4 + 4 - 10 = -2\,.$$

The correct answer is Option A: $$-2$$.