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(C) Let the vertices of the triangle be $O(0), A(\overrightarrow{a}+\overrightarrow{b}),$ and $B(\overrightarrow{a}-\overrightarrow{b})$.The area of t

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$6$

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(C) Let the vertices of the triangle be $O(0), A(\overrightarrow{a}+\overrightarrow{b}),$ and $B(\overrightarrow{a}-\overrightarrow{b})$.The area of the triangle is given by $\frac{1}{2} |\overrightarrow{OA} 	imes \overrightarrow{OB}|$.Substituting the vectors,we get $	ext{Area} = \frac{1}{2} |(\overrightarrow{a}+\overrightarrow{b}) 	imes (\overrightarrow{a}-\overrightarrow{b})|$.Expanding the cross product: $(\overrightarrow{a}+\overrightarrow{b}) 	imes (\overrightarrow{a}-\overrightarrow{b}) = \overrightarrow{a} 	imes \overrightarrow{a} - \overrightarrow{a} 	imes \overrightarrow{b} + \overrightarrow{b} 	imes \overrightarrow{a} - \overrightarrow{b} 	imes \overrightarrow{b}$.Since $\overrightarrow{a} 	imes \overrightarrow{a} = 0$ and $\overrightarrow{b} 	imes \overrightarrow{b} = 0$,and $\overrightarrow{b} 	imes \overrightarrow{a} = -(\overrightarrow{a} 	imes \overrightarrow{b})$,we have:$= 0 - (\overrightarrow{a} 	imes \overrightarrow{b}) - (\overrightarrow{a} 	imes \overrightarrow{b}) - 0 = -2(\overrightarrow{a} 	imes \overrightarrow{b})$.Thus,$	ext{Area} = \frac{1}{2} |-2(\overrightarrow{a} 	imes \overrightarrow{b})| = |\overrightarrow{a} 	imes \overrightarrow{b}|$.Since $\overrightarrow{a} \perp \overrightarrow{b}$,$|\overrightarrow{a} 	imes \overrightarrow{b}| = |\overrightarrow{a}| |\overrightarrow{b}| \sin(90^{\circ}) = 2 	imes 3 	imes 1 = 6$.

If $|\overrightarrow{a}|=2, |\overrightarrow{b}|=3$ and $\overrightarrow{a}, \overrightarrow{b}$ are mutually perpendicular,then the area of the triangle whose vertices are $\overrightarrow{0}, \overrightarrow{a}+\overrightarrow{b}, \overrightarrow{a}-\overrightarrow{b}$ is

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