Let $$\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$$, $$\vec{b} = \hat{i} + \hat{j} + \hat{k}$$ and $$\vec{c}$$ be a vector such that $$\vec{a}  \times (\vec{b}+ \vec{c}) = \vec{0}$$, then the value of $$3(\vec{c} \cdot \vec{a})$$ is equal to ______.

We are given $$\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}, \vec{b} = \hat{i} + \hat{j} + \hat{k}, and a vector \vec{c} such that (\vec{a} + \vec{b}) \times \vec{c} = \vec{0} and \vec{b} \cdot \vec{c} = 5$$.

Adding these vectors, we obtain $$\vec{a} + \vec{b} = (1+1)\hat{i} + (-2+1)\hat{j} + (3+1)\hat{k} = 2\hat{i} - \hat{j} + 4\hat{k}. Since (\vec{a} + \vec{b}) \times \vec{c} = \vec{0} implies that \vec{c} is parallel to \vec{a} + \vec{b}, we set \vec{c} = t(\vec{a} + \vec{b}) = t(2\hat{i} - \hat{j} + 4\hat{k}) for some scalar t$$.

Imposing the condition $$\vec{b} \cdot \vec{c} = 5 gives \vec{b} \cdot \vec{c} = t\,\vec{b} \cdot (\vec{a} + \vec{b}) = t(\vec{b} \cdot \vec{a} + |\vec{b}|^2), where \vec{b} \cdot \vec{a} = 1 - 2 + 3 = 2 and |\vec{b}|^2 = 1 + 1 + 1 = 3. Hence, 5 = t(2 + 3) = 5t, leading to t = 1$$.

Therefore, $$\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}. Computing \vec{c} \cdot \vec{a} yields (2)(1) + (-1)(-2) + (4)(3) = 2 + 2 + 12 = 16. Consequently, 3(\vec{c} \cdot \vec{a}) = 3 \times 16 = 48$$.