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(C) Let $3\vec{a} = \vec{u}$ and $2\vec{b} = \vec{v}$. Given $|\vec{a}| = 2$ and $|\vec{b}| = 3$,we have $|\vec{u}| = 3|\vec{a}| = 6$ and $|\vec{v}| =

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$60$

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(C) Let $3\vec{a} = \vec{u}$ and $2\vec{b} = \vec{v}$. Given $|\vec{a}| = 2$ and $|\vec{b}| = 3$,we have $|\vec{u}| = 3|\vec{a}| = 6$ and $|\vec{v}| = 2|\vec{b}| = 6$.We want to maximize the expression $E = 3|\vec{u} + \vec{v}| + 4|\vec{u} - \vec{v}|$.Let $\alpha$ be the angle between $\vec{u}$ and $\vec{v}$.Using the formula $|\vec{u} \pm \vec{v}| = \sqrt{|\vec{u}|^2 + |\vec{v}|^2 \pm 2|\vec{u}||\vec{v}| \cos \alpha}$,we get:$|\vec{u} + \vec{v}| = \sqrt{6^2 + 6^2 + 2(6)(6) \cos \alpha} = \sqrt{72(1 + \cos \alpha)} = \sqrt{72(2 \cos^2(\alpha/2))} = 12 \cos(\alpha/2)$.$|\vec{u} - \vec{v}| = \sqrt{6^2 + 6^2 - 2(6)(6) \cos \alpha} = \sqrt{72(1 - \cos \alpha)} = \sqrt{72(2 \sin^2(\alpha/2))} = 12 \sin(\alpha/2)$.Substituting these into the expression:$E = 3(12 \cos(\alpha/2)) + 4(12 \sin(\alpha/2)) = 36 \cos(\alpha/2) + 48 \sin(\alpha/2)$.The maximum value of $A \cos x + B \sin x$ is $\sqrt{A^2 + B^2}$.Here,$A = 36$ and $B = 48$,so the maximum value is $\sqrt{36^2 + 48^2} = \sqrt{12^2(3^2 + 4^2)} = 12 \sqrt{9 + 16} = 12(5) = 60$.

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}| = 2$ and $|\vec{b}| = 3$,then the maximum value of $3 |(3\vec{a} + 2\vec{b})| + 4 |(3\vec{a} - 2\vec{b})|$ is:

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