Let $P$ be the plane $\sqrt{3}x+2y+3z=16$ and let
$$S=\{\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}:\alpha^2+\beta^2+\gamma^2=1\text{ and the distance of }(\alpha,\beta,\gamma)\text{ from the plane }P\text{ is }\tfrac{7}{2}\}.$$
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 $\dfrac{80}{\sqrt{3}}V$ is ___.
Step-by-Step Solution
Key Concept: Points in S lie on a small circle of the unit sphere; equilateral triangle scalar triple product via 2·Area·n̂ formula
Distance from $\hat{u}=(\alpha,\beta,\gamma)$ (unit vector) to $P$: $|\sqrt3\alpha+2\beta+3\gamma-16|/4=7/2\Rightarrow|\sqrt3\alpha+2\beta+3\gamma-16|=14$.
Since $|\sqrt3\alpha+2\beta+3\gamma|\leq\sqrt{3+4+9}=4<16$, only the case $\sqrt3\alpha+2\beta+3\gamma=2$ works.
So $S$ = circle on unit sphere where $\hat{n}\cdot\vec{u}=1/2$ with $\hat{n}=(\sqrt3,2,3)/4$. Center of circle: $\vec{c}=\hat{n}/2=(\sqrt3,2,3)/8$, $|\vec{c}|=1/2$. Circle radius: $r=\sqrt{1-1/4}=\sqrt3/2$.
$\vec{u},\vec{v},\vec{w}$ form equilateral triangle on this circle. Let $\vec{u}=\vec{c}+\vec{p}$, etc., with $\vec{p}+\vec{q}+\vec{r}=\vec{0}$, $|\vec{p}|=\sqrt3/2$.
$[\vec{u},\vec{v},\vec{w}]=\vec{c}\cdot(\vec{q}\times\vec{r}+\vec{r}\times\vec{p}+\vec{p}\times\vec{q})$.
$\vec{q}\times\vec{r}+\vec{r}\times\vec{p}+\vec{p}\times\vec{q}=2\cdot\text{Area}(pqr)\cdot\hat{n}$. Side of equilateral triangle inscribed in circle of radius $\sqrt3/2$: $s=\sqrt3\cdot\sqrt3/2=3/2$. Area $=\dfrac{\sqrt3}{4}\cdot\dfrac{9}{4}=\dfrac{9\sqrt3}{16}$.
$\vec{c}\cdot(2\cdot\dfrac{9\sqrt3}{16}\cdot\hat{n})=\dfrac{9\sqrt3}{8}\cdot(\vec{c}\cdot\hat{n})=\dfrac{9\sqrt3}{8}\cdot\dfrac{1}{2}=\dfrac{9\sqrt3}{16}$.
$V=\dfrac{9\sqrt3}{16}$. $\dfrac{80}{\sqrt3}\cdot\dfrac{9\sqrt3}{16}=\dfrac{80\cdot9}{16}=45$.
Correct Answer: 45