If $$\vec{a} = \hat{i} + 2\hat{k}$$, $$\vec{b} = \hat{i} + \hat{j} + \hat{k}$$, $$\vec{c} = 7\hat{i} - 3\hat{j} + 4\hat{k}$$, $$\vec{r} \times \vec{b} + \vec{b} \times \vec{c} = \vec{0}$$ and $$\vec{r} \cdot \vec{a} = 0$$ then $$\vec{r} \cdot \vec{c}$$ is equal to:

We have $$\vec{a} = \hat{i} + 2\hat{k}$$, $$\vec{b} = \hat{i} + \hat{j} + \hat{k}$$, and $$\vec{c} = 7\hat{i} - 3\hat{j} + 4\hat{k}$$.

The conditions are $$\vec{r} \times \vec{b} + \vec{b} \times \vec{c} = \vec{0}$$ and $$\vec{r} \cdot \vec{a} = 0$$.

From $$\vec{r} \times \vec{b} + \vec{b} \times \vec{c} = \vec{0}$$ it follows that $$\vec{r} \times \vec{b} = -\vec{b} \times \vec{c} = \vec{c} \times \vec{b}$$, hence $$(\vec{r} - \vec{c}) \times \vec{b} = \vec{0}$$.

Since $$\vec{r} - \vec{c}$$ is parallel to $$\vec{b}$$, we write $$\vec{r} = \vec{c} + \lambda \vec{b}$$ for some scalar $$\lambda$$.

Substituting gives $$\vec{r} = (7 + \lambda)\hat{i} + (-3 + \lambda)\hat{j} + (4 + \lambda)\hat{k}$$.

Enforcing $$\vec{r} \cdot \vec{a} = 0$$ gives $$(7 + \lambda)(1) + (-3 + \lambda)(0) + (4 + \lambda)(2) = 0$$, which simplifies to $$7 + \lambda + 8 + 2\lambda = 0$$ and hence $$15 + 3\lambda = 0 \implies \lambda = -5$$.

Therefore, $$\vec{r} = 2\hat{i} - 8\hat{j} - \hat{k}$$.

Finally, $$\vec{r} \cdot \vec{c} = 2(7) + (-8)(-3) + (-1)(4) = 14 + 24 - 4 = \boxed{34}$$.

The correct answer is Option A.