Let $P$ be the plane $\sqrt{3}x + 2y + 3z = 16$ and let $S = \{\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k} : \alpha^{2} + \beta^{2} + \gamma^{2} = 1$ and the distance of $(\alpha, \beta, \gamma)$ from the plane $P$ is $\frac{7}{2}\}$. 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 $\frac{80}{\sqrt{3}}V$ is

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

$\text{Plane } P:\quad \sqrt{3}\,x + 2y + 3z = 16$

$S = \{\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k} : \alpha^2+\beta^2+\gamma^2 = 1,\ d\big((\alpha,\beta,\gamma),P\big) = \tfrac{7}{2}\}$

$\vec{u},\vec{v},\vec{w} \in S \text{ distinct},\quad |\vec{u}-\vec{v}| = |\vec{v}-\vec{w}| = |\vec{w}-\vec{u}|$

$\text{To find: } \dfrac{80}{\sqrt{3}}\,V,\qquad V = |[\vec{u}\ \vec{v}\ \vec{w}]|$

$\text{Plan:}\quad \vec{n}\cdot\vec{x} = 2 \Rightarrow S \text{ is a circle}$

$\text{centre } c,\ \text{radius } r$

triangle uvw equilateral ⇒ A; V = 2Ah

$\vec{n} = (\sqrt{3},\ 2,\ 3),\quad |\vec{n}| = \sqrt{3+4+9} = 4$

$d = \dfrac{|\vec{n}\cdot(\alpha,\beta,\gamma) - 16|}{|\vec{n}|} = \dfrac{7}{2}$

$|\vec{n}\cdot\vec{x} - 16| = 14 \Rightarrow \vec{n}\cdot\vec{x} = 30 \text{ or } 2$

$|\vec{n}\cdot\vec{x}| \le |\vec{n}|\,|\vec{x}| = 4 \Rightarrow \vec{n}\cdot\vec{x} = 2$

$S = \{\vec{x} : |\vec{x}| = 1,\ \vec{n}\cdot\vec{x} = 2\} = \text{circle}$

$c = \dfrac{2}{|\vec{n}|^2}\,\vec{n} = \left(\dfrac{\sqrt{3}}{8},\ \dfrac{1}{4},\ \dfrac{3}{8}\right)$

$h = \dfrac{2}{|\vec{n}|} = \dfrac{1}{2},\quad r = \sqrt{1 - h^2} = \dfrac{\sqrt{3}}{2}$

|u⃗-v⃗| = |v⃗-w⃗| = |w⃗-u⃗| ⇒ triangle uvw equilateral

$s = r\sqrt{3} = \dfrac{\sqrt{3}}{2}\cdot\sqrt{3} = \dfrac{3}{2}$

$A = \dfrac{\sqrt{3}}{4}\,s^2 = \dfrac{\sqrt{3}}{4}\cdot\dfrac{9}{4} = \dfrac{9\sqrt{3}}{16}$

V = 6,V_{Ouvw} V_(Ouvw) = 1/3 A h ⇒ V = 6·1/3Ah = 2Ah

$V = 2\cdot\dfrac{9\sqrt{3}}{16}\cdot\dfrac{1}{2} = \dfrac{9\sqrt{3}}{16}$

$\dfrac{80}{\sqrt{3}}\,V = \dfrac{80}{\sqrt{3}}\cdot\dfrac{9\sqrt{3}}{16} = 45$