Question

Let P be the plane √3x+2y+3z =16 and let and let S = {\(\alpha \hat{i}+\beta \hat{j}+\gamma\hat{k}:\alpha^2+\beta^2+\gamma^2=1\) and the distance of (α, β, γ) from the plane P is \(\frac{7}{2}\) }. Let u, v, and 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 \(\vec{u},\vec{v},\vec{w}\). Then the value of \(\frac{80}{\sqrt3}\)V is

Answer 1

Volume of Parallelepiped Determined by Vectors

Let \( P \) be the plane defined by the equation: 

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

Let \( S \) be the set of vectors \( \mathbf{S} = \{\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k} : \alpha^2 + \beta^2 + \gamma^2 = 1 \} \) and the distance of \( (\alpha, \beta, \gamma) \) from the plane \( P \) is given as \( \frac{7}{2} \).

Let \( u, v, \) and \( w \) be three distinct vectors in \( S \) such that:

\[ | \mathbf{u} - \mathbf{v} | = | \mathbf{v} - \mathbf{w} | = | \mathbf{w} - \mathbf{u} | \]

The quantity \( V \) represents the volume of the parallelepiped determined by the vectors \( \mathbf{u}, \mathbf{v}, \mathbf{w} \). The value of \( 80V \) is given as:

\[ \boxed{45} \]

Answer 2

The correct answer will be 45