Let $$\vec{a} = \hat{i} - 3\hat{j} + 7\hat{k}, \vec{b} = 2\hat{i} - \hat{j} + \hat{k}$$ and $$\vec{c}$$ be a vector such that $$(\vec{a} + 2\vec{b}) \times \vec{c} = 3(\vec{c} \times \vec{a})$$. If $$\vec{a} \cdot \vec{c} = 130$$, then $$\vec{b} \cdot \vec{c}$$ is equal to ______

Given $$\vec{a} = \hat{i}-3\hat{j}+7\hat{k}$$, $$\vec{b} = 2\hat{i}-\hat{j}+\hat{k}$$, $$(\vec{a}+2\vec{b})\times\vec{c} = 3(\vec{c}\times\vec{a})$$.

Rewrite: $$(\vec{a}+2\vec{b})\times\vec{c} = -3(\vec{a}\times\vec{c})$$, so $$(\vec{a}+2\vec{b})\times\vec{c} + 3\vec{a}\times\vec{c} = 0$$.

$$(4\vec{a}+2\vec{b})\times\vec{c} = 0$$, meaning $$\vec{c}$$ is parallel to $$4\vec{a}+2\vec{b}$$.

$$4\vec{a}+2\vec{b} = (4+4)\hat{i}+(-12-2)\hat{j}+(28+2)\hat{k} = 8\hat{i}-14\hat{j}+30\hat{k} = 2(4\hat{i}-7\hat{j}+15\hat{k})$$.

So $$\vec{c} = \lambda(4\hat{i}-7\hat{j}+15\hat{k})$$.

$$\vec{a}\cdot\vec{c} = \lambda(4+21+105) = 130\lambda = 130 \Rightarrow \lambda = 1$$.

$$\vec{b}\cdot\vec{c} = 8+7+15 = 30$$.

The answer is 30.