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The volume of the parallelopiped whose edges are represented by $-12 \mathbf{i}+\alpha \mathbf{k}, 3 \mathbf{j}-\mathbf{k}$ and $2 \mathbf{i}+\mathbf{j}-15 \mathbf{k}$ is 546. Then $\alpha=$
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The correct answer is:
-3
Since $\left|\begin{array}{ccc}-12 & 0 & \alpha \\ 0 & 3 & -1 \\ 2 & 1 & -15\end{array}\right|=546 \Rightarrow \alpha=-3$.
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