The volume of the parallelepiped whose edges | vedclass

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(C) The volume of a parallelepiped with edges represented by vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ is given by the absolute value of the scalar tr

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$-3$

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(C) The volume of a parallelepiped with edges represented by vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ is given by the absolute value of the scalar triple product $|\vec{a} \cdot (\vec{b} 	imes \vec{c})|$,which is equal to the determinant of the matrix formed by the components of the vectors.Given vectors are $\vec{a} = -12i + 0j + \alpha k$,$\vec{b} = 0i + 3j - 1k$,and $\vec{c} = 2i + 1j - 15k$.The volume is given by:$|\det(\vec{a}, \vec{b}, \vec{c})| = 546$$\left| \begin{matrix} -12 & 0 & \alpha \ 0 & 3 & -1 \ 2 & 1 & -15 \end{matrix} ight| = \pm 546$Expanding the determinant along the first row:$-12(3(-15) - (-1)(1)) - 0(...) + \alpha(0(1) - 3(2)) = \pm 546$$-12(-45 + 1) + \alpha(-6) = \pm 546$$-12(-44) - 6\alpha = \pm 546$$528 - 6\alpha = \pm 546$Case $1$: $528 - 6\alpha = 546 \Rightarrow -6\alpha = 18 \Rightarrow \alpha = -3$Case $2$: $528 - 6\alpha = -546 \Rightarrow -6\alpha = -1074 \Rightarrow \alpha = 179$Comparing with the given options,the correct value is $\alpha = -3$.

The volume of the parallelepiped whose edges are represented by $-12i + \alpha k$,$3j - k$,and $2i + j - 15k$ is $546$. Then $\alpha = $

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