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The lines are given by $\vec{r} = \vec{a_1} + \lambda\vec{b}$ and $\vec{r} = \vec{a_2} + \mu\vec{b}$, where $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$. The distance $d$ between these parallel lines is given by: $$d = \frac{|\vec{b} \times (\vec{a_2} - \vec{a_1})|}{|\vec{b}|}$$ Here, $\vec{a_2} - \vec{a_1} = (3-1)\hat{i} + (3-2)\hat{j} + (-5 - (-4))\hat{k} = 2\hat{i} + \hat{j} - \hat{k}$.
$\vec{b} \times (\vec{a_2} - \vec{a_1}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 2 & 1 & -1 \end{vmatrix} = \hat{i}(-3-6) - \hat{j}(-2-12) + \hat{k}(2-6) = -9\hat{i} + 14\hat{j} - 4\hat{k}$. The magnitude is $\sqrt{(-9)^2 + 14^2 + (-4)^2} = \sqrt{81 + 196 + 16} = \sqrt{293}$. The magnitude of $\vec{b}$ is $\sqrt{2^2 + 3^2 + 6^2} = \sqrt{4+9+36} = 7$. Thus, $d = \frac{\sqrt{293}}{7}$.
Since $d$ represents the side length of the square, the area is $d^2 = \frac{293}{49}$ square units.
Final Answer: 293/49
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