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The vector $\vec{AB}$ is given by the difference of the position vectors of points B and A.
$\vec{AB} = \vec{OB} - \vec{OA} = (8\hat{i} - \hat{j} + 0\hat{k}) - (2\hat{i} + \hat{j} + 3\hat{k}) = (8-2)\hat{i} + (-1-1)\hat{j} + (0-3)\hat{k} = 6\hat{i} - 2\hat{j} - 3\hat{k}$
The vector in the opposite direction of $\vec{AB}$ is $-\vec{AB}$.
$-\vec{AB} = - (6\hat{i} - 2\hat{j} - 3\hat{k}) = -6\hat{i} + 2\hat{j} + 3\hat{k}$
The magnitude of $-\vec{AB}$ is given by:
$|-\vec{AB}| = \sqrt{(-6)^2 + (2)^2 + (3)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$
The unit vector in the direction of $-\vec{AB}$ is given by:
$\hat{u} = \frac{-\vec{AB}}{|-\vec{AB}|} = \frac{-6\hat{i} + 2\hat{j} + 3\hat{k}}{7} = -\frac{6}{7}\hat{i} + \frac{2}{7}\hat{j} + \frac{3}{7}\hat{k}$
The vector of magnitude 21 in the direction of $-\vec{AB}$ is given by:
$\vec{v} = 21 \hat{u} = 21 \left( -\frac{6}{7}\hat{i} + \frac{2}{7}\hat{j} + \frac{3}{7}\hat{k} \right) = -18\hat{i} + 6\hat{j} + 9\hat{k}$
Final Answer: $-18\hat{i} + 6\hat{j} + 9\hat{k}$