Let three points A(2,3,4), B(3,4,2) and C(4,2 | vedclass

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(B) First,calculate the side lengths of the triangle $ABC$: $AB = \sqrt{(3-2)^2 + (4-3)^2 + (2-4)^2} = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6}$. $BC = \s

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$\sqrt{3}$

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(B) First,calculate the side lengths of the triangle $ABC$: $AB = \sqrt{(3-2)^2 + (4-3)^2 + (2-4)^2} = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6}$. $BC = \sqrt{(4-3)^2 + (2-4)^2 + (3-2)^2} = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$. $CA = \sqrt{(2-4)^2 + (3-2)^2 + (4-3)^2} = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{6}$. Since $AB = BC = CA = \sqrt{6}$,the triangle $ABC$ is an equilateral triangle with side length $a = \sqrt{6}$. Point $D$ is at a distance of $\sqrt{6}$ from $A, B,$ and $C$. This means $DA = DB = DC = \sqrt{6}$. Thus,$ABCD$ is a regular tetrahedron with all edges equal to $\sqrt{6}$. The volume $V$ of a regular tetrahedron with edge length $a$ is given by $V = \frac{a^3}{6\sqrt{2}}$. Substituting $a = \sqrt{6}$: $V = \frac{(\sqrt{6})^3}{6\sqrt{2}} = \frac{6\sqrt{6}}{6\sqrt{2}} = \sqrt{\frac{6}{2}} = \sqrt{3}$ cubic units.

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

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