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(B) The area of a triangle with vertices $A$,$B$,and $C$ is given by $\Delta = \frac{1}{2} |\overrightarrow{AB} 	imes \overrightarrow{AC}|$.First,fin

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$2\sqrt{6}$

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(B) The area of a triangle with vertices $A$,$B$,and $C$ is given by $\Delta = \frac{1}{2} |\overrightarrow{AB} 	imes \overrightarrow{AC}|$.First,find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$:$\overrightarrow{AB} = (2-1)i + (0-(-1))j + (-1-2)k = i + j - 3k$$\overrightarrow{AC} = (0-1)i + (2-(-1))j + (1-2)k = -i + 3j - k$Next,calculate the cross product $\overrightarrow{AB} 	imes \overrightarrow{AC}$:$\overrightarrow{AB} 	imes \overrightarrow{AC} = \begin{vmatrix} i & j & k \ 1 & 1 & -3 \ -1 & 3 & -1 \end{vmatrix}$$= i(-1 - (-9)) - j(-1 - 3) + k(3 - (-1))$$= i(8) - j(-4) + k(4) = 8i + 4j + 4k$Now,find the magnitude of the cross product:$|\overrightarrow{AB} 	imes \overrightarrow{AC}| = \sqrt{8^2 + 4^2 + 4^2} = \sqrt{64 + 16 + 16} = \sqrt{96} = 4\sqrt{6}$Finally,the area of the triangle is:$\Delta = \frac{1}{2} |\overrightarrow{AB} 	imes \overrightarrow{AC}| = \frac{1}{2} (4\sqrt{6}) = 2\sqrt{6}$.

If the vertices of $\Delta ABC$ are $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$,then what is the area of the triangle?

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