To find the volume of a tetrahedron given vertices $A(5, 4, -3)$, $B(-6, -6, -5)$, $C(4, -6, -4)$, and $D(0, 0, 10)$, we can use the following formula:

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

Step-by-Step Solution

1. Find Vectors $\vec{AB}$, $\vec{AC}$, and $\vec{AD}$:

   $\vec{AB} = (-6 - 5, -6 - 4, -5 + 3) = (-11, -10, -2)$ $\vec{AC} = (4 - 5, -6 - 4, -4 + 3) = (-1, -10, -1)$ $\vec{AD} = (0 - 5, 0 - 4, 10 + 3) = (-5, -4, 13)$

2. Compute the Cross Product $\vec{AC} \times \vec{AD}$:

   Using the determinant for the cross product: $\vec{AC} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -10 & -1 \\ -5 & -4 & 13 \end{vmatrix}$

   Expanding the determinant: $\vec{AC} \times \vec{AD} = \hat{i}( (-10)(13) - (-1)(-4) ) - \hat{j}( (-1)(13) - (-1)(-5) ) + \hat{k}( (-1)(-4) - (-10)(-5) )$

   Simplifying each component: $= \hat{i}(-130 - 4) - \hat{j}(-13 - 5) + \hat{k}(4 - 50)$ $= \hat{i}(-134) + \hat{j}(18) - \hat{k}(46)$ So, $\vec{AC} \times \vec{AD} = (-134, 18, -46)$

3. Calculate the Dot Product $\vec{AB} \cdot (\vec{AC} \times \vec{AD})$:

   $\vec{AB} \cdot (\vec{AC} \times \vec{AD}) = (-11)(-134) + (-10)(18) + (-2)(-46)$ $= 1474 - 180 + 92 = 1386$

4. Calculate the Volume:

   $V = \frac{1}{6} |1386| = \frac{1386}{6} = 231$

Final Answer

The volume of the tetrahedron is: $\boxed{231}$

Would you like further details or have any questions?

1. How is the cross product of two vectors calculated in 3D?
2. What are some other methods to find the volume of a tetrahedron in 3D?
3. How does the determinant help in calculating vector products?
4. Why is the factor $\frac{1}{6}$ used in the volume formula for a tetrahedron?
5. Can this method be applied to any shape other than a tetrahedron?

Tip: Always double-check calculations, especially signs, when dealing with vector products, as mistakes can lead to incorrect volume values.