Let $$\vec{a} = 6\hat{i} + 9\hat{j} + 12\hat{k}$$, $$\vec{b} = \alpha\hat{i} + 11\hat{j} - 2\hat{k}$$ and $$\vec{c}$$ be vectors such that $$\vec{a} \times \vec{c} = \vec{a} \times \vec{b}$$. If $$\vec{a} \cdot \vec{c} = -12$$, and $$\vec{c} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 5$$ then $$\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k})$$ is equal to ______.

We have vectors $$\vec{a} = 6\hat{i} + 9\hat{j} + 12\hat{k}$$, $$\vec{b} = \alpha\hat{i} + 11\hat{j} - 2\hat{k}$$, and the condition $$\vec{a} \times \vec{c} = \vec{a} \times \vec{b}$$.

Since $$\vec{a} \times \vec{c} = \vec{a} \times \vec{b}$$ implies $$\vec{a} \times (\vec{c} - \vec{b}) = \vec{0}$$, the vector $$\vec{c} - \vec{b}$$ must be parallel to $$\vec{a}$$. Thus we write $$\vec{c} = \vec{b} + \lambda\vec{a}$$ for some scalar $$\lambda$$, which in components gives $$\vec{c} = (\alpha + 6\lambda)\hat{i} + (11 + 9\lambda)\hat{j} + (-2 + 12\lambda)\hat{k}$$.

Applying the condition $$\vec{c} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 5$$ leads to $$(\alpha + 6\lambda) - 2(11 + 9\lambda) + (-2 + 12\lambda) = 5$$, which simplifies to $$\alpha + 6\lambda - 22 - 18\lambda - 2 + 12\lambda = 5$$ and hence to $$\alpha - 24 = 5\implies \alpha = 29$$.

Next, using $$\vec{a} \cdot \vec{c} = -12$$ and substituting $$\alpha = 29$$ into the expression for $$\vec{c}$$ gives $$6(29 + 6\lambda) + 9(11 + 9\lambda) + 12(-2 + 12\lambda) = -12$$. This expands to $$174 + 36\lambda + 99 + 81\lambda - 24 + 144\lambda = -12$$, so $$249 + 261\lambda = -12$$, yielding $$\lambda = -1$$.

Substituting $$\alpha = 29$$ and $$\lambda = -1$$ into $$\vec{c} = (\alpha + 6\lambda)\hat{i} + (11 + 9\lambda)\hat{j} + (-2 + 12\lambda)\hat{k}$$ gives $$\vec{c} = 23\hat{i} + 2\hat{j} - 14\hat{k}$$.

Finally, the required dot product is $$\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 23 + 2 + (-14) = 11$$, so the answer is $$11$$.