Let P be the plane $\sqrt{3}\mathrm{x}+2\mathrm{y}+3\mathrm{z}=16$ and let $\mathrm{S}=\left\{\mathrm{\alpha }\hat{\mathrm{i}}+\mathrm{\beta }\hat{\mathrm{j}}+\mathrm{\gamma }\hat{\mathrm{k}}: {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+{\mathrm{\gamma }}^2=1\right.$and the distance of $(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma })$ from the plane P  is $\frac{7}{2}\}.$ Let $\overrightarrow{\mathrm{u}}, \overrightarrow{\mathrm{v}} \mathrm{and} \overrightarrow{\mathrm{w}}$ and  be three distinct vectors in S such that $|\overrightarrow{\mathrm{u}}-\overrightarrow{\mathrm{v}}|=|\overrightarrow{\mathrm{v}}-\overrightarrow{\mathrm{w}}|=|\overrightarrow{\mathrm{w}}-\overrightarrow{\mathrm{u}}|$. Let V be the volume of the parallelepiped determined by vectors $\overrightarrow{\mathrm{u}}, \overrightarrow{\mathrm{v}} \mathrm{and} \overrightarrow{\mathrm{w}}$. Then the value of $\frac{80}{\sqrt{3}}$V is

Let $\mathrm{P}:\sqrt{3}\mathrm{x}+2\mathrm{y}+3\mathrm{z}=16$
Let P₁ bet the plane which is at a distance of $\frac{7}{2}$ units from P. 
Let $\mathrm{A}\left(\overline{\mathrm{u}}\right),\mathrm{B}\left(\overline{\mathrm{v}}\right),\mathrm{C}\left(\overline{\mathrm{w}}\right)$ are 3 points on P₁ such that $\left|\overline{\mathrm{u}}\right|=\left|\overline{\mathrm{v}}\right|=\left|\overline{\mathrm{w}}\right|=1$ and $|\overline{\mathrm{u}}-\overline{\mathrm{v}}|=|\overline{\mathrm{v}}-\overline{\mathrm{w}}|=|\overline{\mathrm{w}}-\overline{\mathrm{u}}|$
Let P and Q are the projections drawn from origin ‘O’ to the planes P and  respectively. 
$\begin{array}{l}\therefore \mathrm{OP}=4,\mathrm{QP}=\frac{7}{2}\Rightarrow \mathrm{OQ}=\frac{1}{2}\text{ and }\mathrm{OA}=\mathrm{OB}=\mathrm{OC}=1\\ \therefore \mathrm{AQ}=\mathrm{BQ}=\mathrm{CQ}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}\\ \Rightarrow \mathrm{AB}=\sqrt{\frac{3}{4}+\frac{3}{4}-2\cdot \frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}\left(\frac{-1}{2}\right)}\end{array}$
$\therefore$$∆$ABC is equilateral triangle and Q be its circumcentre) 
$\begin{array}{l}\left|\begin{array}{ccc}1& \frac{-1}{8}& \frac{-1}{8}\\ \frac{-1}{8}& 1& \frac{-1}{8}\\ \frac{-1}{8}& \frac{-1}{8}& 1\end{array}\right|=\left(1-\frac{1}{64}\right)+\frac{1}{8}\left(-\frac{1}{8}-\frac{1}{64}\right)-\frac{1}{8}\left(\frac{1}{64}+\frac{1}{8}\right)\\ =\left(1-\frac{1}{64}\right)-\frac{2}{8}\left(\frac{1}{8}+\frac{1}{64}\right)=\frac{243}{256}\Rightarrow \mathrm{V}=\left|[\stackrel{ }{\overline{\mathrm{u}} \overline{\mathrm{v}} } \overline{\mathrm{w}}]\right|=\frac{9\sqrt{3}}{16}\Rightarrow \frac{80\mathrm{V}}{\sqrt{3}}=45\\ \end{array}$