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(C) Let the adjacent sides of the parallelogram be vectors $\vec{a} = i - 2j + 3k$ and $\vec{b} = 2i + j - 4k$.The area of a parallelogram with adjace

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$5\sqrt{6}$

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(C) Let the adjacent sides of the parallelogram be vectors $\vec{a} = i - 2j + 3k$ and $\vec{b} = 2i + j - 4k$.The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product: $	ext{Area} = |\vec{a} 	imes \vec{b}|$.First,calculate the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} i & j & k \ 1 & -2 & 3 \ 2 & 1 & -4 \end{vmatrix}$$= i((-2)(-4) - (3)(1)) - j((1)(-4) - (3)(2)) + k((1)(1) - (-2)(2))$$= i(8 - 3) - j(-4 - 6) + k(1 + 4)$$= 5i + 10j + 5k$.Now,calculate the magnitude of the resulting vector:$|\vec{a} 	imes \vec{b}| = \sqrt{5^2 + 10^2 + 5^2}$$= \sqrt{25 + 100 + 25}$$= \sqrt{150}$$= \sqrt{25 	imes 6} = 5\sqrt{6}$.Thus,the area is $5\sqrt{6}$ square units.

The area of a parallelogram whose adjacent sides are $i - 2j + 3k$ and $2i + j - 4k$ is:

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