AMC 10

Let P be the plane √3x + 2y + 3z = 16 and let S = {αî + βĵ + γk̂ : α² + β² + γ² = 1 and the distance of (α, β, γ) from the plane P is 7/2}. Let u⃗, v⃗ and w⃗ be three distinct vectors in S such that |u⃗ - v⃗| = |v⃗ - w⃗| = |w⃗ - u⃗|. Let V be the volume of the parallelepiped determined by vectors u⃗, v⃗ and w⃗. Then the value of 80/√3V is $P:\ \sqrt{3}x+2y+3z=16$ $S= \{\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k} : \alpha^{2}+\beta^{2}+\gamma^{2}=1,\ d((\alpha,\beta,\gamma),P)=\tfrac{7}{2}\}$ $|\vec{u}-\vec{v}|=|\vec{v}-\vec{w}|=|\vec{w}-\vec{u}|$ $\text{Find } \dfrac{80}{\sqrt{3}}V, \quad V=[\vec{u}\ \vec{v}\ \vec{w}]$ $1.\ d(\vec{r},P)=\dfrac{7}{2} \Rightarrow \vec{n}\cdot\vec{r}$ $2.\ S \text{ is a circle: centre, radius}$ $3.\ V=|\vec{u}\cdot(\vec{v}\times\vec{w})|, \ \text{find } \dfrac{80}{\sqrt{3}}V$ $|\vec{r}|=1: \ \alpha^{2}+\beta^{2}+\gamma^{2}=1$ $\vec{n}=(\sqrt{3},2,3), \quad |\vec{n}|=\sqrt{3+4+9}=4$ $d=\dfrac{|\vec{n}\cdot\vec{r}-16|}{4}=\dfrac{7}{2}$ $|\vec{n}\cdot\vec{r}|\le |\vec{n}|\,|\vec{r}|=4 \Rightarrow \vec{n}\cdot\vec{r}-16<0$ $16-\vec{n}\cdot\vec{r}=14 \Rightarrow \vec{n}\cdot\vec{r}=2$ $S: |\vec{r}|=1, \ \vec{n}\cdot\vec{r}=2$ $O_{S}=\dfrac{2}{|\vec{n}|^{2}}\vec{n}=\left(\dfrac{\sqrt{3}}{8},\dfrac{1}{4},\dfrac{3}{8}\right), \quad |O_{S}|=\dfrac{1}{2}$ $\rho=\sqrt{1-\dfrac{1}{4}}=\dfrac{\sqrt{3}}{2}$ $u,v,w \in S, \quad |u-v|=|v-w|=|w-u|$ $s=\rho\sqrt{3}=\dfrac{\sqrt{3}}{2}\cdot\sqrt{3}=\dfrac{3}{2}$ $V=|\vec{u}\cdot((\vec{v}-\vec{u})\times(\vec{w}-\vec{u}))|$ $(\vec{v}-\vec{u})\times(\vec{w}-\vec{u}) = 2(\text{Area})\,\hat{n}$ $\hat{n}\cdot\vec{u}=\dfrac{2}{4}=\dfrac{1}{2}$ $V=2(\text{Area})\cdot\dfrac{1}{2}=\text{Area}$ $\text{Area}=\dfrac{\sqrt{3}}{4}s^{2}=\dfrac{\sqrt{3}}{4}\cdot\dfrac{9}{4}$ $V=\dfrac{9\sqrt{3}}{16}$ $\dfrac{80}{\sqrt{3}}V=\dfrac{80}{\sqrt{3}}\cdot\dfrac{9\sqrt{3}}{16}$ $\boxed{45}$