Let $$\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$$ and $$\vec{b} = 7\hat{i} + \hat{j} - 6\hat{k}$$. If $$\vec{r} \times \vec{a} = \vec{r} \times \vec{b}$$, $$\vec{r} \cdot (\hat{i} + 2\hat{j} + \hat{k}) = -3$$, then $$\vec{r} \cdot (2\hat{i} - 3\hat{j} + \hat{k})$$ is equal to:

We are given $$\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$$ and $$\vec{b} = 7\hat{i} + \hat{j} - 6\hat{k}$$, with the conditions $$\vec{r} \times \vec{a} = \vec{r} \times \vec{b}$$ and $$\vec{r} \cdot (\hat{i} + 2\hat{j} + \hat{k}) = -3$$.

From $$\vec{r} \times \vec{a} = \vec{r} \times \vec{b}$$, we get $$\vec{r} \times (\vec{a} - \vec{b}) = \vec{0}$$.

Now $$\vec{a} - \vec{b} = (2-7)\hat{i} + (-3-1)\hat{j} + (4+6)\hat{k} = -5\hat{i} - 4\hat{j} + 10\hat{k}$$.

Since $$\vec{r} \times (\vec{a} - \vec{b}) = \vec{0}$$, the vector $$\vec{r}$$ is parallel to $$\vec{a} - \vec{b}$$. So $$\vec{r} = \lambda(-5\hat{i} - 4\hat{j} + 10\hat{k})$$ for some scalar $$\lambda$$.

Using $$\vec{r} \cdot (\hat{i} + 2\hat{j} + \hat{k}) = -3$$: $$\lambda(-5 - 8 + 10) = -3$$, so $$-3\lambda = -3$$, giving $$\lambda = 1$$.

Therefore $$\vec{r} = -5\hat{i} - 4\hat{j} + 10\hat{k}$$.

Now $$\vec{r} \cdot (2\hat{i} - 3\hat{j} + \hat{k}) = (-5)(2) + (-4)(-3) + (10)(1) = -10 + 12 + 10 = 12$$.

The answer is Option A: 12.