Volume of Parallelepiped Formed by Vectors on Distance Constraint Plane

To find the value of $\frac{80}{\sqrt{3}} V$, we first analyze the set $S$ of vectors $\vec{r} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$.

Step 1: Geometry of the Set $S$

The vectors in $S$ are unit vectors, which means their terminal points lie on the unit sphere: $\alpha^2 + \beta^2 + \gamma^2 = 1 \implies |\vec{r}| = 1$

The distance $d$ of a point $(\alpha, \beta, \gamma)$ from the plane $P: \sqrt{3}x + 2y + 3z = 16$ is given as $\frac{7}{2}$. Using the perpendicular distance formula: $d = \frac{|\sqrt{3}\alpha + 2\beta + 3\gamma - 16|}{\sqrt{(\sqrt{3})^2 + 2^2 + 3^2}} = \frac{|\sqrt{3}\alpha + 2\beta + 3\gamma - 16|}{4} = \frac{7}{2}$

This yields: $|\sqrt{3}\alpha + 2\beta + 3\gamma - 16| = 14$

So, we have two possible cases:

1. $\sqrt{3}\alpha + 2\beta + 3\gamma = 30$
2. $\sqrt{3}\alpha + 2\beta + 3\gamma = 2$

By the Cauchy-Schwarz Inequality, for any point on the unit sphere $\alpha^2 + \beta^2 + \gamma^2 = 1$: $|\sqrt{3}\alpha + 2\beta + 3\gamma| \le \sqrt{(\sqrt{3})^2 + 2^2 + 3^2} \sqrt{\alpha^2 + \beta^2 + \gamma^2} = 4 \cdot 1 = 4$

Hence, $\sqrt{3}\alpha + 2\beta + 3\gamma = 30$ is not possible. Therefore, we must have: $\sqrt{3}\alpha + 2\beta + 3\gamma = 2$

If we define the unit normal to the plane $P$ as $\hat{n} = \frac{\sqrt{3}\hat{i} + 2\hat{j} + 3\hat{k}}{4}$, the condition becomes: $\vec{r} \cdot \hat{n} = \frac{2}{4} = \frac{1}{2}$

Thus, the set $S$ represents a circle $C$ formed by the intersection of the unit sphere $|\vec{r}| = 1$ and the plane $\vec{r} \cdot \hat{n} = \frac{1}{2}$.

Step 2: Radius of Circle $C$ and Distance between Vectors

The distance from the origin to the plane of the circle $C$ is $h = \frac{1}{2}$. The radius $R$ of the circle $C$ is: $R = \sqrt{1^2 - h^2} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \frac{\sqrt{3}}{2}$

The vectors $\vec{u}, \vec{v}, \vec{w} \in S$ form an equilateral triangle inscribed in this circle $C$, as their mutual distances are equal: $|\vec{u} - \vec{v}| = |\vec{v} - \vec{w}| = |\vec{w} - \vec{u}| = a$

The side length $a$ of an equilateral triangle inscribed in a circle of radius $R$ is: $a = R\sqrt{3} = \frac{\sqrt{3}}{2} \times \sqrt{3} = \frac{3}{2}$

Step 3: Determining the Scalar Triple Product (Volume $V$)

To find the volume $V = |[\vec{u} \quad \vec{v} \quad \vec{w}]|$ of the parallelepiped, we use the Gram determinant.

First, we find the dot products between the vectors: Since $|\vec{u}| = |\vec{v}| = |\vec{w}| = 1$: $|\vec{u} - \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2(\vec{u} \cdot \vec{v}) = \left(\frac{3}{2}\right)^2$ $1 + 1 - 2(\vec{u} \cdot \vec{v}) = \frac{9}{4} \implies 2(\vec{u} \cdot \vec{v}) = -\frac{1}{4} \implies \vec{u} \cdot \vec{v} = -\frac{1}{8}$

By symmetry: $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{w} = \vec{w} \cdot \vec{u} = -\frac{1}{8}$

The square of the scalar triple product is given by the determinant of the Gram matrix: $V^2 = [\vec{u} \quad \vec{v} \quad \vec{w}]^2 = \begin{vmatrix} \vec{u} \cdot \vec{u} & \vec{u} \cdot \vec{v} & \vec{u} \cdot \vec{w} \\ \vec{v} \cdot \vec{u} & \vec{v} \cdot \vec{v} & \vec{v} \cdot \vec{w} \\ \vec{w} \cdot \vec{u} & \vec{w} \cdot \vec{v} & \vec{w} \cdot \vec{w} \end{vmatrix} = \begin{vmatrix} 1 & -1/8 & -1/8 \\ -1/8 & 1 & -1/8 \\ -1/8 & -1/8 & 1 \end{vmatrix}$

Evaluating the determinant: $V^2 = 1\left(1 - \frac{1}{64}\right) - \left(-\frac{1}{8}\right)\left(-\frac{1}{8} - \frac{1}{64}\right) + \left(-\frac{1}{8}\right)\left(\frac{1}{64} - \left(-\frac{1}{8}\right)\right)$ $V^2 = \frac{63}{64} - \frac{9}{512} - \frac{9}{512} = \frac{63}{64} - \frac{18}{512} = \frac{504 - 18}{512} = \frac{486}{512} = \frac{243}{256}$

Taking the square root to find $V$: $V = \sqrt{\frac{243}{256}} = \frac{9\sqrt{3}}{16}$

Step 4: Final Calculation

We need to calculate $\frac{80}{\sqrt{3}}V$: $\frac{80}{\sqrt{3}} V = \frac{80}{\sqrt{3}} \times \frac{9\sqrt{3}}{16} = 5 \times 9 = 45$