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(C) Let the vertices of the tetrahedron be $A = (1, -1, -2)$,$B = (-2, 1, -2)$,$C = (-1, -2, 1)$,and $D = (2, 2, a)$.The volume of a tetrahedron with

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-$1$

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(C) Let the vertices of the tetrahedron be $A = (1, -1, -2)$,$B = (-2, 1, -2)$,$C = (-1, -2, 1)$,and $D = (2, 2, a)$.The volume of a tetrahedron with vertices $A, B, C, D$ is given by $V = \frac{1}{6} |(\vec{b}-\vec{a}) \cdot ((\vec{c}-\vec{a}) 	imes (\vec{d}-\vec{a}))|$.First,calculate the vectors:$\vec{AB} = \vec{b}-\vec{a} = (-2-1)\hat{i} + (1-(-1))\hat{j} + (-2-(-2))\hat{k} = -3\hat{i} + 2\hat{j} + 0\hat{k}$$\vec{AC} = \vec{c}-\vec{a} = (-1-1)\hat{i} + (-2-(-1))\hat{j} + (1-(-2))\hat{k} = -2\hat{i} - 1\hat{j} + 3\hat{k}$$\vec{AD} = \vec{d}-\vec{a} = (2-1)\hat{i} + (2-(-1))\hat{j} + (a-(-2))\hat{k} = 1\hat{i} + 3\hat{j} + (a+2)\hat{k}$The volume is $\frac{1}{6} |\det(\vec{AB}, \vec{AC}, \vec{AD})| = \frac{20}{3}$,so $|\det(\vec{AB}, \vec{AC}, \vec{AD})| = 40$.Calculate the determinant:$\det = \begin{vmatrix} -3 & 2 & 0 \ -2 & -1 & 3 \ 1 & 3 & a+2 \end{vmatrix} = -3(-1(a+2) - 9) - 2(-2(a+2) - 3) + 0 = -3(-a-11) - 2(-2a-7) = 3a + 33 + 4a + 14 = 7a + 47$.Setting $|7a + 47| = 40$:Case $1$: $7a + 47 = 40 \implies 7a = -7 \implies a = -1$.Case $2$: $7a + 47 = -40 \implies 7a = -87 \implies a = -87/7$ (not an integer).Thus,the integral value of $a$ is $-1$.

Let the volume of the tetrahedron with vertices $\hat{i}-\hat{j}-2\hat{k}$,$-2\hat{i}+\hat{j}-2\hat{k}$,$-\hat{i}-2\hat{j}+\hat{k}$,and $2\hat{i}+2\hat{j}+a\hat{k}$ be $\frac{20}{3}$. Then the integral value of $a$ is

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