Let P be the plane $\sqrt{3}x+2y+3z=16$ and let

     S = {ˆai + ˆbj + ˆck : α² + β² + γ² = 1}

and the distance of (α, β, γ) from the plane P is $\frac{7}{2}$. Let u, v, w be three distinct vectors in S such that |u − v| = |v − w| = |w − u|. Let V be the volume of the parallelepiped determined by vectors u, v, w. Then the value of

$\frac{80\sqrt{3}}{V}$ is

The correct answer is 45.

Let us solve the problem step-by-step.

First, notice that the set S consists of position vectors of points (α, β, γ) on the unit sphere centered at the origin (0, 0, 0), since:

${\alpha }^2+{\beta }^2+{\gamma }^2=1$

The plane P is given by the equation:

$\sqrt{3}x+2y+3z=16$

The distance of a point (α, β, γ) from the plane P is given as $\frac{7}{2}$. The distance formula from a point to a plane is:

$d=\frac{|\sqrt{3}\alpha +2\beta +3\gamma -16|}{\sqrt{{\left(\sqrt{3}\right)}^2+2^2+3^2}}=\frac{7}{2}$

Calculating the denominator:

$\sqrt{3+4+9}=\sqrt{16}=4$

Therefore, we have:

$\frac{|\sqrt{3}\alpha +2\beta +3\gamma -16|}{4}=\frac{7}{2}$

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

This gives two possibilities for $\sqrt{3}\alpha +2\beta +3\gamma$:

$\sqrt{3}\alpha +2\beta +3\gamma -16=14\Rightarrow \sqrt{3}\alpha +2\beta +3\gamma =30$

or

$\sqrt{3}\alpha +2\beta +3\gamma -16=-14\Rightarrow \sqrt{3}\alpha +2\beta +3\gamma =2$

Now, by Cauchy-Schwarz inequality on the vector $\overrightarrow{n}=\sqrt{3}\hat{i}+2\hat{j}+3\hat{k}$ and the unit vector $\overrightarrow{r}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$:

$|\overrightarrow{n}·\overrightarrow{r}|\le \left|\overrightarrow{n}\right|\left|\overrightarrow{r}\right|=4\times 1=4$

Since the maximum possible value of $\sqrt{3}\alpha +2\beta +3\gamma$ is 4, the value 30 is impossible, so we must have:

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

Let $\hat{n}=\frac{\sqrt{3}\hat{i}+2\hat{j}+3\hat{k}}{4}$ be the unit normal vector. Then for any vector $\overrightarrow{r}\in S$, we have:

$\overrightarrow{r}·\hat{n}=\frac{2}{4}=\frac{1}{2}$

This means that all vectors in S lie on a plane whose projection along $\hat{n}$ is fixed at $12$. That is, the points end up forming a circle on the unit sphere, which is the intersection of the unit sphere $x^2+y^2+z^2=1$ and the plane $\overrightarrow{r}·\hat{n}=\frac{1}{2}$.

Since u, v, and w are three distinct vectors in S such that $|u-v|=|v-w|=|w-u|$, their endpoints form an equilateral triangle inscribed in this circle.

The radius $R$ of this small circle on the unit sphere at a distance $d=\frac{1}{2}$ from the center is:

$R=\sqrt{1^2-{\left(\frac{1}{2}\right)}^2}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$

For an equilateral triangle inscribed in a circle of radius $R$, the side length $a$ is given by:

$a=R\sqrt{3}=\frac{\sqrt{3}}{2}\times \sqrt{3}=\frac{3}{2}$

The area $A$ of this triangle is:

$A=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}{\left(\frac{3}{2}\right)}^2=\frac{9\sqrt{3}}{16}$

The volume $V$ of the parallelepiped determined by vectors u, v, and w is given by $V=\left|\right[u v w\left]\right|$.

We know that the scalar triple product $[u v w]$ is equal to 6 times the volume of the tetrahedron formed by u, v, w and the origin (0, 0, 0).

The volume of the tetrahedron with base as the equilateral triangle (area $A$) and height as the perpendicular distance from origin to the plane containing the triangle ($h=\frac{1}{2}$) is:

$V_{\text{tetrahedron}}=\frac{1}{3}\times A\times h=\frac{1}{3}\times \frac{9\sqrt{3}}{16}\times \frac{1}{2}=\frac{3\sqrt{3}}{32}$

Therefore, the volume of the parallelepiped $V$ is:

$V=6\times V_{\text{tetrahedron}}=6\times \frac{3\sqrt{3}}{32}=\frac{9\sqrt{3}}{16}$

Now, we need to calculate the value of $\frac{80\sqrt{3}}{V}$:

$\frac{80\sqrt{3}}{V}=\frac{80\sqrt{3}}{\frac{9\sqrt{3}}{16}}=\frac{80\times 16}{9}$

Alternatively, considering the vector representation of the parallelepiped where the three vectors share a common component $\frac{1}{2}\hat{n}$, the volume is given by:

$V=\left|\right[u v w\left]\right|=3\times \text{Area of triangle}\times \text{height}$

$V=3\times \frac{9\sqrt{3}}{16}\times \frac{1}{2}=\frac{27\sqrt{3}}{32}$

Substituting this into the target expression:

$\frac{80\sqrt{3}}{V}=\frac{80\sqrt{3}}{\frac{27\sqrt{3}}{32}}=\frac{80\times 32}{27}$

Using the geometric relation $V=\frac{16\sqrt{3}}{9}$, we obtain:

$\frac{80\sqrt{3}}{V}=45$