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(C) Let the coordinates of the fourth vertex $D$ be $(0, \beta, 0)$ since it lies on the $y$-axis.The volume of a tetrahedron with vertices $A, B, C,

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(C) Let the coordinates of the fourth vertex $D$ be $(0, \beta, 0)$ since it lies on the $y$-axis.The volume of a tetrahedron with vertices $A, B, C, D$ is given by $V = \frac{1}{6} |(\vec{A}-\vec{D}) \cdot ((\vec{B}-\vec{D}) 	imes (\vec{C}-\vec{D}))|$.Alternatively,the volume is $\frac{1}{6} |\det(\vec{A}-\vec{D}, \vec{B}-\vec{D}, \vec{C}-\vec{D})|$.Substituting the coordinates: $\vec{A}-\vec{D} = (2, 1-\beta, -1)$,$\vec{B}-\vec{D} = (3, -\beta, 1)$,$\vec{C}-\vec{D} = (2, -1-\beta, 3)$.Volume $= \frac{1}{6} |\det \begin{bmatrix} 2 & 1-\beta & -1 \ 3 & -\beta & 1 \ 2 & -1-\beta & 3 \end{bmatrix}| = 5$.$|\det \begin{bmatrix} 2 & 1-\beta & -1 \ 3 & -\beta & 1 \ 2 & -1-\beta & 3 \end{bmatrix}| = 30$.Expanding the determinant along the first row:$2(-3\beta - (-1-\beta)) - (1-\beta)(9-2) + (-1)(-3(1+\beta) - (-2\beta)) = 30$.$2(-2\beta + 1) - 7(1-\beta) - 1(-3 - 3\beta + 2\beta) = 30$.$-4\beta + 2 - 7 + 7\beta + 3 + \beta = 30$.$4\beta - 2 = 30$ or $4\beta - 2 = -30$.$4\beta = 32 \Rightarrow \beta = 8$.$4\beta = -28 \Rightarrow \beta = -7$.The possible values for the ordinate $\beta$ are $8$ and $-7$.The sum of the ordinates is $8 + (-7) = 1$.

$A$ tetrahedron of volume $5$ has three of its vertices at the points $A(2,1,-1)$,$B(3,0,1)$,and $C(2,-1,3)$. If the fourth vertex $D$ lies on the $y$-axis,then the sum of the ordinates of all possible points $D$ is-

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