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Let $\bar{a}=4 \bar{i}+5 \bar{j}-\bar{k}, \bar{b}=\bar{i}-4 \bar{j}+5 \bar{k}, \bar{c}=3 \bar{i}+\bar{j}-\bar{k}$ and let $\bar{\alpha}$ be a vector perpendicular to both $\bar{a}$ and $\bar{b}$ such that $\bar{\alpha} \cdot \bar{c}=63$. Then $\bar{\alpha}=$
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The correct answer is:
$21 \bar{i}-21 \bar{j}-21 \bar{k}$
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