Let P be the plane sqrt{3} x+2 y+3 z=16 and l | vedclass

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(B) Given $|\vec{u}-\vec{v}|=|\vec{v}-\vec{w}|=|\vec{w}-\vec{u}|$,so $	riangle UVW$ is an equilateral triangle.Let $O$ be the origin. The vectors $\v

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$45$

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(B) Given $|\vec{u}-\vec{v}|=|\vec{v}-\vec{w}|=|\vec{w}-\vec{u}|$,so $	riangle UVW$ is an equilateral triangle.Let $O$ be the origin. The vectors $\vec{u}, \vec{v}, \vec{w}$ are unit vectors,so they lie on a sphere of radius $1$ centered at $O$.The distance of the plane $P: \sqrt{3}x + 2y + 3z = 16$ from the origin $O(0,0,0)$ is $OQ = \frac{|0+0+0-16|}{\sqrt{(\sqrt{3})^2 + 2^2 + 3^2}} = \frac{16}{\sqrt{3+4+9}} = \frac{16}{4} = 4$.The distance of any point $(\alpha, \beta, \gamma) \in S$ from the plane $P$ is given as $\frac{7}{2}$. Let $Q$ be the projection of $O$ on $P$. Since $U, V, W$ are at a constant distance from $P$,they lie on a circle which is the intersection of the sphere and a plane parallel to $P$. Let $P'$ be this plane. The distance of $P'$ from $O$ is $OP = OQ - PQ = 4 - \frac{7}{2} = \frac{1}{2}$.The radius of the circle formed by the intersection of the sphere and plane $P'$ is $R = \sqrt{1^2 - (1/2)^2} = \sqrt{3/4} = \frac{\sqrt{3}}{2}$.Since $	riangle UVW$ is equilateral and inscribed in this circle of radius $R$,its side length $a = 2R \cos(30^\circ) = 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3}{2}$.The area of $	riangle UVW = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \cdot \frac{9}{4} = \frac{9\sqrt{3}}{16}$.The volume of the tetrahedron with vertices $O, U, V, W$ is $\frac{1}{3} 	imes 	ext{Area}(	riangle UVW) 	imes OP = \frac{1}{3} 	imes \frac{9\sqrt{3}}{16} 	imes \frac{1}{2} = \frac{3\sqrt{3}}{32}$.The volume $V$ of the parallelepiped determined by $\vec{u}, \vec{v}, \vec{w}$ is $6 	imes 	ext{Volume of tetrahedron} = 6 	imes \frac{3\sqrt{3}}{32} = \frac{9\sqrt{3}}{16}$.Therefore,$\frac{80}{\sqrt{3}} V = \frac{80}{\sqrt{3}} 	imes \frac{9\sqrt{3}}{16} = 5 	imes 9 = 45$.

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