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(A) Let the cube $Q$ have vertices $(x_1, x_2, x_3)$ where $x_i \in \{0, 1\}$.Consider the main diagonal $OG$ connecting $(0,0,0)$ and $(1,1,1)$. Its

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$\frac{1}{\sqrt{6}}$

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(A) Let the cube $Q$ have vertices $(x_1, x_2, x_3)$ where $x_i \in \{0, 1\}$.Consider the main diagonal $OG$ connecting $(0,0,0)$ and $(1,1,1)$. Its direction vector is $\vec{v}_1 = (1, 1, 1)$. The equation of line $OG$ is $\frac{x}{1} = \frac{y}{1} = \frac{z}{1}$.Consider a face diagonal,for example,the diagonal $AB$ on the face $z=0$ connecting $(1,0,0)$ and $(0,1,0)$. Its direction vector is $\vec{v}_2 = (-1, 1, 0)$.The shortest distance $d$ between two lines with direction vectors $\vec{v}_1, \vec{v}_2$ and points $P_1, P_2$ is given by $d = \frac{|(\vec{P}_2 - \vec{P}_1) \cdot (\vec{v}_1 	imes \vec{v}_2)|}{|\vec{v}_1 	imes \vec{v}_2|}$.Here $\vec{v}_1 	imes \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ -1 & 1 & 0 \end{vmatrix} = \hat{i}(0-1) - \hat{j}(0 - (-1)) + \hat{k}(1 - (-1)) = -\hat{i} - \hat{j} + 2\hat{k}$.The magnitude is $|\vec{v}_1 	imes \vec{v}_2| = \sqrt{(-1)^2 + (-1)^2 + 2^2} = \sqrt{6}$.Taking $P_1 = (0,0,0)$ and $P_2 = (1,0,0)$,$\vec{P}_2 - \vec{P}_1 = (1,0,0)$.The distance is $d = \frac{|(1,0,0) \cdot (-1, -1, 2)|}{\sqrt{6}} = \frac{|-1|}{\sqrt{6}} = \frac{1}{\sqrt{6}}$.For other combinations of face diagonals and main diagonals,the distance is either $0$ (if they intersect) or $\frac{1}{\sqrt{6}}$. Thus,the maximum distance is $\frac{1}{\sqrt{6}}$.

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