The problem asks us to find the volume of a tetrahedron with vertices $A(2, -1, -3)$, $B(4, 1, 3)$, $C(3, 2, -1)$, and $D(1, 4, 2)$.

1. Problem Description

The problem asks us to find the volume of a tetrahedron with vertices $A(2, -1, -3)$, $B(4, 1, 3)$, $C(3, 2, -1)$, and $D(1, 4, 2)$.

2. Solution Steps

The volume $V$ of a tetrahedron with vertices $A$, $B$, $C$, and $D$ can be calculated using the scalar triple product of the vectors $\vec{AB}$, $\vec{AC}$, and $\vec{AD}$:

$V = \frac{1}{6} |(\vec{AB} \times \vec{AC}) \cdot \vec{AD}|$

First, find the vectors $\vec{AB}$, $\vec{AC}$, and $\vec{AD}$:

$\vec{AB} = B - A = (4-2, 1-(-1), 3-(-3)) = (2, 2, 6)$

$\vec{AC} = C - A = (3-2, 2-(-1), -1-(-3)) = (1, 3, 2)$

$\vec{AD} = D - A = (1-2, 4-(-1), 2-(-3)) = (-1, 5, 5)$

Next, find the cross product $\vec{AB} \times \vec{AC}$:

$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 6 \\ 1 & 3 & 2 \end{vmatrix} = \hat{i}(2 \cdot 2 - 6 \cdot 3) - \hat{j}(2 \cdot 2 - 6 \cdot 1) + \hat{k}(2 \cdot 3 - 2 \cdot 1) = \hat{i}(4 - 18) - \hat{j}(4 - 6) + \hat{k}(6 - 2) = -14\hat{i} + 2\hat{j} + 4\hat{k} = (-14, 2, 4)$

Now, find the scalar triple product $(\vec{AB} \times \vec{AC}) \cdot \vec{AD}$:

$(\vec{AB} \times \vec{AC}) \cdot \vec{AD} = (-14, 2, 4) \cdot (-1, 5, 5) = (-14)(-1) + (2)(5) + (4)(5) = 14 + 10 + 20 = 44$

Finally, find the volume $V$:

$V = \frac{1}{6} |(\vec{AB} \times \vec{AC}) \cdot \vec{AD}| = \frac{1}{6} |44| = \frac{44}{6} = \frac{22}{3} = 7\frac{1}{3}$

3. Final Answer

$7\frac{1}{3}$