The vector equation of the plane passing through the intersection of the planes $$\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$$ and $$\vec{r} \cdot (\hat{i} - 2\hat{j}) = -2$$, and the point (1, 0, 2) is:

The two given planes are $$\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$$ and $$\vec{r} \cdot (\hat{i} - 2\hat{j}) = -2$$. The family of planes passing through their intersection is given by $$\vec{r} \cdot [(\hat{i} + \hat{j} + \hat{k}) + \lambda(\hat{i} - 2\hat{j})] = 1 + \lambda(-2)$$, which simplifies to $$\vec{r} \cdot [(1 + \lambda)\hat{i} + (1 - 2\lambda)\hat{j} + \hat{k}] = 1 - 2\lambda$$.

Since this plane passes through the point $$(1, 0, 2)$$, we substitute to get $$(1 + \lambda)(1) + (1 - 2\lambda)(0) + (1)(2) = 1 - 2\lambda$$, which gives $$1 + \lambda + 2 = 1 - 2\lambda$$, so $$3\lambda = -2$$ and $$\lambda = -\frac{2}{3}$$.

Substituting $$\lambda = -\frac{2}{3}$$ back into the equation, the normal vector becomes $$\left(1 - \frac{2}{3}\right)\hat{i} + \left(1 + \frac{4}{3}\right)\hat{j} + \hat{k} = \frac{1}{3}\hat{i} + \frac{7}{3}\hat{j} + \hat{k}$$, and the right-hand side becomes $$1 + \frac{4}{3} = \frac{7}{3}$$.

Multiplying the entire equation by 3, we obtain $$\vec{r} \cdot (\hat{i} + 7\hat{j} + 3\hat{k}) = 7$$.

The correct answer is $$\vec{r} \cdot (\hat{i} + 7\hat{j} + 3\hat{k}) = 7$$.