In tetrahedron $A B C D,$ the face $A B C$ is a regular (Equilateral triangle) and the face $B C D$ is perpendicular to it.
$∠ D A C = \frac{π}{3}, | A D | = 6 u n i t s .$
Angle between the lines $\vec{A D} a n d \vec{B C} i s \cos^{- 1} \frac{1}{4}$ and if ‘A’ origin,
$\vec{A D} = \vec{d}, \vec{A B} = \vec{b}, \vec{A C} = \vec{c} .$
Volume of tetrahedron $A B C D$ (in Cu. Units) is

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$\frac{9}{4}$

$\frac{27}{8}$

$\frac{27}{4}$

$\frac{8}{3}$

answer is B.

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Detailed Solution

Let $α = \cos^{- 1} \frac{1}{4} \Rightarrow \cos α = \frac{1}{4}$

$\frac{\vec{A D} . \vec{B C}}{| \vec{A D} | | \vec{B C} |} = \frac{\vec{d} . (\vec{c} - \vec{b})}{| \vec{d} | | \vec{c} - \vec{b} |} \Rightarrow 4 \vec{d} . (\vec{c} - \vec{b}) = | \vec{d} | | \vec{c} - \vec{b} | = | \vec{b} | | \vec{d} |$

Since $| \vec{b} - \vec{c} | = | \vec{b} | = | \vec{c} |$
Let angle between $\vec{d}$ and $\vec{b}$ is $θ .$ And the angle between $\vec{d}$ and $\vec{c}$ is $\frac{π}{3} .$

$\Rightarrow 4 (| \vec{d} | | \vec{c} | \frac{1}{2} - | \vec{d} | | \vec{b} | \cos θ) = | \vec{d} | | \vec{b} | \Rightarrow 4 (\frac{1}{2} - \cos θ) = 1 \Rightarrow \cos θ = \frac{1}{4} A B C ⊥ D B C \Rightarrow (\vec{A B} \times \vec{A C}) . (\vec{B D} \times \vec{D C}) = 0 \Rightarrow (\vec{b} \times \vec{c}) . ((\vec{b} - \vec{d}) \times (\vec{c} - \vec{d})) = 0 \Rightarrow (\vec{b} \times \vec{c}) . (\vec{b} \times \vec{c} - \vec{b} \times \vec{d} - \vec{d} \times \vec{c}) = 0 \Rightarrow {| \vec{b} \times \vec{c} |}^{2} - (\vec{b} \times \vec{c}) . (\vec{b} \times \vec{d}) - (\vec{b} \times \vec{c}) . (\vec{d} \times \vec{c}) = 0 \Rightarrow {| \vec{b} |}^{2} {| \vec{c} |}^{2} \frac{3}{4} - | \begin{matrix} \vec{b} . \vec{b} & \vec{b} . \vec{d} \\ \vec{c} . \vec{b} & \vec{c} . \vec{d} \end{matrix} | - | \begin{matrix} \vec{b} . \vec{d} & \vec{b} . \vec{c} \\ \vec{c} . \vec{d} & \vec{c} . \vec{c} \end{matrix} | = 0 \Rightarrow \frac{3 | \vec{b} |}{4} - 6 (\frac{1}{2} - \frac{1}{8}) - 6 (\frac{1}{4} - \frac{1}{4}) = 0 \Rightarrow | \vec{b} | = 3 a n d | \vec{c} | = 3 \Rightarrow | \vec{b} | + | \vec{c} | = 6 {[\begin{matrix} \vec{d} & \vec{b} & \vec{c} \end{matrix}]}^{2} = | \begin{matrix} \vec{d} . \vec{d} & \vec{d} . \vec{b} & \vec{d} . \vec{c} \\ \vec{b} . \vec{d} & \vec{b} . \vec{b} & \vec{b} . \vec{c} \\ \vec{c} . \vec{d} & \vec{c} . \vec{b} & \vec{c} . \vec{c} \end{matrix} | = {| \vec{d} |}^{2} {| \vec{b} |}^{2} {| \vec{c} |}^{2} | \begin{matrix} 1 & \frac{1}{4} & \frac{1}{2} \\ \frac{1}{4} & 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 1 \end{matrix} |$