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Volume of Parallelopiped

Let three poi...

Question

Let three points $A (2, 3, 4), B (3, 4, 2)$ and $C (4, 2, 3)$ in space are given. A point $D$ in space is such that it is at a distance of $\sqrt{6}$ units from the three given points. The volume of tetrahedron $A B C D$ is -

A

1

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B

\sqrt{3}

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C

\sqrt{13}

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D

2

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Solution

The correct option is B $\sqrt{3}$
$A B = B C = C A =$ $\sqrt{6}$

It is a regular tetrahedron

$[\to a \to b \to c]^{2} = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \to a . \to a & \to a . \to b & \to a . \to c \to b . \to a & \to b . \to b & \to b . \to c \to c . \to a & \to c . \to b & \to c . \to c \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} 6 & 3 & 3 3 & 6 & 3 3 & 3 & 6 \end{matrix} ∣ ∣ ∣ ∣$

$= 108$

Volume $= \frac{1}{6} [\to a \to b \to c]$ $= \frac{1}{6} \sqrt{108} = \sqrt{3}$

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Volume of Parallelopiped

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