Let P be the plane $\sqrt{3} x + 2 y + 3 z = 16$ and let $S = {α \hat{i} + β \hat{j} + γ \hat{k}; α^{2} + β^{2} + γ^{2} = 1 a n d t h e distance of (α, β, γ) f r o m t h e p l a n e P i s \frac{7}{2}}$ . Let $\vec{u}, \vec{v} a n d \vec{ω}$ be three distinct vectors in S such that $| \vec{u} - \vec{v} | = | \vec{v} - \vec{ω} | = | \vec{ω} - \vec{u} |$ . Let V be the volume of the parallelopiped determined by vectors $\vec{u}, \vec{v}$ and $\vec{ω}$ .Then the value of V is_____

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$\frac{3}{16}$

$\frac{9}{16}$

$\frac{9 \sqrt{3}}{16}$

$\frac{3 \sqrt{3}}{16}$

answer is A.

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Detailed Solution

OE=distance b/w (0,0,0) and plane =4
OD= $4 - \frac{7}{2} = \frac{1}{2}$
$Δ A B C$ is equilateral
In $Δ O D B, B D = \sqrt{O B^{2} - O D^{2}} = \sqrt{1 - \frac{1}{2}} - \sqrt{\frac{3}{2}}$ $F o r Δ A B C, a = 2 R \sin A \Rightarrow a = 2. \frac{\sqrt{3}}{2} . \frac{\sqrt{3}}{2} \Rightarrow a = \frac{3}{2} \Rightarrow A r (Δ A B C) = \frac{\sqrt{3}}{4} . \frac{9}{4} = \frac{9 \sqrt{3}}{16}$