Let $\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}$, $\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a} \times \vec{c} = \vec{b}$. If $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$, then $|\vec{a} + \vec{c}|^2$ is equal to:

$\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}$

$\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}$

$\vec{c} = c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k}$

$\vec{a} \times \vec{c} = \vec{b}$

$(2c_3 - 2c_2)\hat{i} + (c_3 + 2c_1)\hat{j} - (c_2 + 2c_1)\hat{k} = 8\hat{i} + 7\hat{j} - 3\hat{k}$

$2c_3 - 2c_2 = 8,\quad c_3 + 2c_1 = 7,\quad c_2 + 2c_1 = 3$

$c_1 + c_2 + c_3 = 4,\quad c_1 + c_2 + c_3 = 4$

$c_1 = 2,\quad c_2 = -1,\quad c_3 = 3$

$|\vec{a} + \vec{c}|^2 = | \hat{i} + \hat{j} + 5\hat{k} |^2 = 27$