Let P be the plane $$\sqrt{3}x + 2y + 3z = 16$$ and let $$S = \left\{\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k} : \alpha^2 + \beta^2 + \gamma^2 = 1 \text{ and the distance of } (\alpha, \beta, \gamma) \text{ from the plane P is } \frac{7}{2}\right\}$$. Let $$\vec{u}$$, $$\vec{v}$$ and $$\vec{w}$$ be three distinct vectors in S such that $$|\vec{u} - \vec{v}| = |\vec{v} - \vec{w}| = |\vec{w} - \vec{u}|$$. Let V be the volume of the parallelepiped determined by vectors $$\vec{u}$$, $$\vec{v}$$ and $$\vec{w}$$. Then the value of $$\frac{80}{\sqrt{3}}V$$ is

The given plane is $$\sqrt{3}\,x+2y+3z=16$$.
For any point $$(\alpha,\beta,\gamma)$$ the (perpendicular) distance from this plane equals

$$\frac{\left|\sqrt{3}\alpha+2\beta+3\gamma-16\right|} {\sqrt{(\sqrt{3})^{2}+2^{2}+3^{2}}} \;=\;\frac{\left|\sqrt{3}\alpha+2\beta+3\gamma-16\right|}{4}.$$

We are told that this distance is $$\dfrac72$$, hence

$$\left|\sqrt{3}\alpha+2\beta+3\gamma-16\right|=14.$$

The left-hand side can never reach $$30$$ for a unit vector because
$$\max\bigl(\sqrt{3}\alpha+2\beta+3\gamma\bigr)=\sqrt{(\sqrt{3})^{2}+2^{2}+3^{2}}=4.$$
Therefore only the negative sign is possible and

$$\sqrt{3}\alpha+2\beta+3\gamma=2.$$ Thus $$S$$ is the circle obtained by the intersection of

• the unit sphere $$\alpha^{2}+\beta^{2}+\gamma^{2}=1,$$ and
• the plane $$\sqrt{3}x+2y+3z=2.$$

Let $$\vec{n}=(\sqrt{3},2,3),\;|\vec{n}|=4,\; \hat{n}=\dfrac{\vec{n}}{|\vec{n}|}.$$
The plane is at (signed) distance $$d=\frac{2}{4}=\frac12$$ from the origin along $$\hat{n}$$, so the circle has

centre $$\vec{C}=d\,\hat{n}=\frac12\hat{n},$$
radius $$r=\sqrt{1-d^{2}}=\sqrt{1-\frac14}=\frac{\sqrt3}{2}.$$

Choose two unit vectors $$\vec{e}_{1},\vec{e}_{2}$$ lying in the plane and orthogonal to each other such that $$\vec{e}_{1}\times\vec{e}_{2}=\hat{n}.$$

The three required vectors can now be written as vertices of an equilateral triangle of radius $$r$$ inside this circle:

$$\begin{aligned} \vec{u}&=\vec{C}+r\,\vec{e}_{1},\\ \vec{v}&=\vec{C}+r\Bigl(-\tfrac12\vec{e}_{1}+\tfrac{\sqrt3}{2}\vec{e}_{2}\Bigr),\\ \vec{w}&=\vec{C}+r\Bigl(-\tfrac12\vec{e}_{1}-\tfrac{\sqrt3}{2}\vec{e}_{2}\Bigr). \end{aligned}$$

Define
$$\vec{A}_{1}=\vec{e}_{1},\quad \vec{A}_{2}=-\tfrac12\vec{e}_{1}+\tfrac{\sqrt3}{2}\vec{e}_{2},\quad \vec{A}_{3}=-\tfrac12\vec{e}_{1}-\tfrac{\sqrt3}{2}\vec{e}_{2}.$$

Then $$\vec{u}=\vec{C}+r\vec{A}_{1},\; \vec{v}=\vec{C}+r\vec{A}_{2},\; \vec{w}=\vec{C}+r\vec{A}_{3}.$$

The volume of the parallelepiped is the absolute value of the scalar triple product

$$V=\left|\vec{u}\,\cdot\,(\vec{v}\times\vec{w})\right|.$$

Subtract the first column from the next two to simplify the determinant:

$$\bigl(\vec{u},\vec{v},\vec{w}\bigr)= \det\!\bigl[\vec{C}+r\vec{A}_{1},\; r(\vec{A}_{2}-\vec{A}_{1}),\; r(\vec{A}_{3}-\vec{A}_{1})\bigr]$$ $$=r^{2}\det\!\bigl[\vec{C}+r\vec{A}_{1},\; \vec{B}_{2},\;\vec{B}_{3}\bigr],$$ where $$\vec{B}_{2}=\vec{A}_{2}-\vec{A}_{1} =-\tfrac32\vec{e}_{1}+\tfrac{\sqrt3}{2}\vec{e}_{2},\quad \vec{B}_{3}=\vec{A}_{3}-\vec{A}_{1} =-\tfrac32\vec{e}_{1}-\tfrac{\sqrt3}{2}\vec{e}_{2}.$$

Because $$\vec{A}_{1},\vec{B}_{2},\vec{B}_{3}$$ all lie in the plane (they are perpendicular to $$\hat{n}$$),
$$\det\bigl[\vec{A}_{1},\vec{B}_{2},\vec{B}_{3}\bigr]=0,$$ so only $$\vec{C}$$ contributes:

$$\vec{u}\cdot(\vec{v}\times\vec{w})= r^{2}\det\!\bigl[\vec{C},\vec{B}_{2},\vec{B}_{3}\bigr] =r^{2}\,\vec{C}\cdot(\vec{B}_{2}\times\vec{B}_{3}).$$

The cross product is computed using $$\vec{e}_{1}\times\vec{e}_{2}=\hat{n}:$$ $$\vec{B}_{2}\times\vec{B}_{3} =\left(-\tfrac32\vec{e}_{1}+\tfrac{\sqrt3}{2}\vec{e}_{2}\right) \times \left(-\tfrac32\vec{e}_{1}-\tfrac{\sqrt3}{2}\vec{e}_{2}\right) =\frac{3\sqrt3}{2}\,\hat{n}.$$

Therefore

$$\begin{aligned} \vec{C}\cdot(\vec{B}_{2}\times\vec{B}_{3}) &=\frac12\hat{n}\cdot\frac{3\sqrt3}{2}\hat{n} =\frac{3\sqrt3}{4},\\[6pt] r^{2} &=\left(\frac{\sqrt3}{2}\right)^{2}=\frac34,\\[6pt] V &=\left|\frac34\cdot\frac{3\sqrt3}{4}\right| =\frac{9\sqrt3}{16}. \end{aligned}$$

Finally,

$$\frac{80}{\sqrt3}V =\frac{80}{\sqrt3}\cdot\frac{9\sqrt3}{16} =\frac{720}{16}=45.$$

Hence the required value is 45.