If the volume of tetrahedron $A B C D$ is $1$ cubic units, where $A (0, 1, 2), B (- 1, 2, 1)$ and $C (1, 2, 1)$ , then the locus of point $D$ is-

x + y - z = 3

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y + z = 6

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y + z = 0

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y + z = - 3

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Solution

The correct options are
A $y + z = 0$
C $y + z = 6$
Volume of tetrahedron $A B C D$ is $\frac{1}{6} ∣ ∣ [\to A B \to A C \to A D] ∣ ∣ = 1$ cubic units.
$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} - 1 & 1 & - 1 1 & 1 & 1 x - 0 & y - 0 & z - 2 \end{matrix} ∣ ∣ ∣ ∣ = \pm 6$
$\Rightarrow - 2 (y - 1) - 2 (z - 2) = \pm 6$
$\Rightarrow y + z - 3 = \pm 3$
$\Rightarrow y + z = 6$ or $y + z = 0$

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If the volume of tetrahedron ABCD is 1 cubic units, where A(0,1,2),B(−1,2,1) and C(1,2,1), then the locus of point D is-

The correct options are A y+z=0 C y+z=6Volume of tetrahedron ABCD is 16∣∣[→AB →AC →AD]∣∣=1cubic units.⇒∣∣ ∣∣−11−1111x−0y−0z−2∣∣ ∣∣=±6⇒−2(y−1)−2(z−2)=±6⇒y+z−3=±3⇒y+z=6 or y+z=0

If the volume of tetrahedron $A B C D$ is $1$ cubic units, where $A (0, 1, 2), B (- 1, 2, 1)$ and $C (1, 2, 1)$ , then the locus of point $D$ is-