Let $$\vec{a} = 2\hat{i} + 5\hat{j} - \hat{k}$$, $$\vec{b} = 2\hat{i} - 2\hat{j} + 2\hat{k}$$ and $$\vec{c}$$ be three vectors such that $$(\vec{c} + \hat{i}) \times (\vec{a} + \vec{b} + \hat{i}) = \vec{a} \times (\vec{c} + \hat{i})$$. If $$\vec{a} \cdot \vec{c} = -29$$, then $$\vec{c} \cdot (-2\hat{i} + \hat{j} + \hat{k})$$ is equal to:

Given $$\vec{a} = 2\hat{i} + 5\hat{j} - \hat{k}$$, $$\vec{b} = 2\hat{i} - 2\hat{j} + 2\hat{k}$$, and:

$$ (\vec{c} + \hat{i}) \times (\vec{a} + \vec{b} + \hat{i}) = \vec{a} \times (\vec{c} + \hat{i}) $$

Let $$\vec{d} = \vec{c} + \hat{i}$$. Using $$\vec{a} \times \vec{d} = -\vec{d} \times \vec{a}$$:

$$ \vec{d} \times (\vec{a} + \vec{b} + \hat{i}) = -\vec{d} \times \vec{a} $$ $$ \vec{d} \times (2\vec{a} + \vec{b} + \hat{i}) = \vec{0} $$

So $$\vec{d}$$ is parallel to $$2\vec{a} + \vec{b} + \hat{i}$$.

$$2\vec{a} + \vec{b} + \hat{i} = 7\hat{i} + 8\hat{j}$$.

So $$\vec{d} = \lambda(7\hat{i} + 8\hat{j})$$ and $$\vec{c} = (7\lambda - 1)\hat{i} + 8\lambda\hat{j}$$.

Using $$\vec{a} \cdot \vec{c} = -29$$:

$$ 2(7\lambda - 1) + 5(8\lambda) = -29 \implies 54\lambda - 2 = -29 \implies \lambda = -\frac{1}{2} $$

$$\vec{c} = -\frac{9}{2}\hat{i} - 4\hat{j}$$.

$$\vec{c} \cdot (-2\hat{i} + \hat{j} + \hat{k}) = 9 - 4 + 0 = 5$$.

The correct answer is Option 4: 5.