Let $$\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$$, $$\vec{b} = \hat{i} - \hat{j}$$ and $$\vec{c} = \hat{i} - \hat{j} - \hat{k}$$ be three given vectors. If $$\vec{r}$$ is a vector such that $$\vec{r} \times \vec{a} = \vec{c} \times \vec{a}$$ and $$\vec{r} \cdot \vec{b} = 0$$, then $$\vec{r} \cdot \vec{a}$$ is equal to ______

We are given $$\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$$, $$\vec{b} = \hat{i} - \hat{j}$$, $$\vec{c} = \hat{i} - \hat{j} - \hat{k}$$, and the conditions $$\vec{r} \times \vec{a} = \vec{c} \times \vec{a}$$ and $$\vec{r} \cdot \vec{b} = 0$$.

From $$\vec{r} \times \vec{a} = \vec{c} \times \vec{a}$$, we get $$(\vec{r} - \vec{c}) \times \vec{a} = \vec{0}$$. This means $$\vec{r} - \vec{c}$$ is parallel to $$\vec{a}$$, so $$\vec{r} = \vec{c} + \lambda\vec{a}$$ for some scalar $$\lambda$$.

Substituting: $$\vec{r} = (1 + \lambda)\hat{i} + (-1 + 2\lambda)\hat{j} + (-1 - \lambda)\hat{k}$$.

Applying the condition $$\vec{r} \cdot \vec{b} = 0$$ where $$\vec{b} = \hat{i} - \hat{j}$$: $$(1 + \lambda)(1) + (-1 + 2\lambda)(-1) + (-1 - \lambda)(0) = 0$$.

This gives $$1 + \lambda + 1 - 2\lambda = 0$$, so $$2 - \lambda = 0$$, hence $$\lambda = 2$$.

Therefore, $$\vec{r} = 3\hat{i} + 3\hat{j} - 3\hat{k}$$.

Computing $$\vec{r} \cdot \vec{a} = 3(1) + 3(2) + (-3)(-1) = 3 + 6 + 3 = 12$$.