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(D) We know that $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b}) + 2(\vec{c} \cdot \vec{a}) + 2(\vec

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$3\sqrt{6}$

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(D) We know that $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b}) + 2(\vec{c} \cdot \vec{a}) + 2(\vec{b} \cdot \vec{c})$.Since $\vec{c}$ is perpendicular to the plane of $\vec{a}$ and $\vec{b}$, we have $\vec{c} \cdot \vec{a} = 0$ and $\vec{c} \cdot \vec{b} = 0$.Also, $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos(\frac{\pi}{4}) = 3 	imes 2\sqrt{2} 	imes \frac{1}{\sqrt{2}} = 6$.Substituting these values:$|\vec{a}+\vec{b}+\vec{c}|^2 = 3^2 + (2\sqrt{2})^2 + 5^2 + 2(6) + 0 + 0$$|\vec{a}+\vec{b}+\vec{c}|^2 = 9 + 8 + 25 + 12 = 54$.Therefore, $|\vec{a}+\vec{b}+\vec{c}| = \sqrt{54} = 3\sqrt{6}$.

Let $\vec{a}, \vec{b}, \vec{c}$ be $3$ vectors such that $|\vec{a}|=3, |\vec{b}|=2\sqrt{2}, |\vec{c}|=5$ and $\vec{c}$ is perpendicular to the plane of $\vec{a}$ and $\vec{b}$. If the angle between the vectors $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$, then $|\vec{a}+\vec{b}+\vec{c}|=$

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