Let triangle PQR be a triangle. Let vec{a}=ov | vedclass

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(C) In $	riangle PQR$,we have $\vec{a} + \vec{b} + \vec{c} = 0$,which implies $\vec{b} + \vec{c} = -\vec{a}$.Taking the magnitude on both sides,$|\ve

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$(A, C, D)$

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(C) In $	riangle PQR$,we have $\vec{a} + \vec{b} + \vec{c} = 0$,which implies $\vec{b} + \vec{c} = -\vec{a}$.Taking the magnitude on both sides,$|\vec{b} + \vec{c}|^2 = |-\vec{a}|^2 = |\vec{a}|^2$.$|\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{b} \cdot \vec{c}) = |\vec{a}|^2$.Substituting the given values: $(4\sqrt{3})^2 + |\vec{c}|^2 + 2(24) = 12^2$.$48 + |\vec{c}|^2 + 48 = 144 \Rightarrow |\vec{c}|^2 = 48 \Rightarrow |\vec{c}| = 4\sqrt{3}$.Now,check option $(A)$: $\frac{|\vec{c}|^2}{2} - |\vec{a}| = \frac{48}{2} - 12 = 24 - 12 = 12$. Thus,$(A)$ is true.Check option $(B)$: $\frac{|\vec{c}|^2}{2} + |\vec{a}| = 24 + 12 = 36 
eq 30$. Thus,$(B)$ is false.For option $(D)$,since $\vec{a} + \vec{b} = -\vec{c}$,we have $|\vec{a} + \vec{b}|^2 = |-\vec{c}|^2 = |\vec{c}|^2$.$|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{c}|^2$.$144 + 48 + 2(\vec{a} \cdot \vec{b}) = 48 \Rightarrow 2(\vec{a} \cdot \vec{b}) = -144 \Rightarrow \vec{a} \cdot \vec{b} = -72$. Thus,$(D)$ is true.For option $(C)$,since $\vec{a} + \vec{b} + \vec{c} = 0$,we have $\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a}$.$|\vec{a} 	imes \vec{b} + \vec{c} 	imes \vec{a}| = |\vec{a} 	imes \vec{b} + \vec{a} 	imes \vec{b}| = 2|\vec{a} 	imes \vec{b}|$.$|\vec{a} 	imes \vec{b}| = |\vec{a}||\vec{b}| \sin(	heta)$,where $	heta$ is the angle between $\vec{a}$ and $\vec{b}$.$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos(	heta) = -72 \Rightarrow 12(4\sqrt{3}) \cos(	heta) = -72 \Rightarrow 48\sqrt{3} \cos(	heta) = -72 \Rightarrow \cos(	heta) = -\frac{72}{48\sqrt{3}} = -\frac{\sqrt{3}}{2}$.So,$\sin(	heta) = \sqrt{1 - (-\frac{\sqrt{3}}{2})^2} = \frac{1}{2}$.$|\vec{a} 	imes \vec{b}| = 12(4\sqrt{3})(\frac{1}{2}) = 24\sqrt{3}$.$2|\vec{a} 	imes \vec{b}| = 2(24\sqrt{3}) = 48\sqrt{3}$. Thus,$(C)$ is true.Therefore,$(A, C, D)$ are true.

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