The volume of the tetrahedron whose vertices are `A(1,-1,10),B(-1,-3,7),C(5,-1,1) and D(7,-4,7)` is

Options

A. 26
B. 29
C. 32
D. None of these

Answer

To find the volume of the tetrahedron with vertices \( A(1, -1, 10) \), \( B(-1, -3, 7) \), \( C(5, -1, 1) \), and \( D(7, -4, 7) \), we will use the formula for the volume of a tetrahedron given by:

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

### Step 1: Find the vectors \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \)

1. **Calculate \( \vec{AB} \)**:
\[ \vec{AB} = B - A = (-1 - 1, -3 + 1, 7 - 10) = (-2, -2, -3) \]

2. **Calculate \( \vec{AC} \)**:
\[ \vec{AC} = C - A = (5 - 1, -1 + 1, 1 - 10) = (4, 0, -9) \]

3. **Calculate \( \vec{AD} \)**:
\[ \vec{AD} = D - A = (7 - 1, -4 + 1, 7 - 10) = (6, -3, -3) \]

### Step 2: Compute the scalar triple product \( \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \)

1. **Calculate the cross product \( \vec{AC} \times \vec{AD} \)**:
\[ \vec{AC} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 0 & -9 \\ 6 & -3 & -3 \end{vmatrix} \]

Expanding the determinant:
\[ = \hat{i} \begin{vmatrix} 0 & -9 \\ -3 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & -9 \\ 6 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & 0 \\ 6 & -3 \end{vmatrix} \]

Calculating the 2x2 determinants:
\[ = \hat{i} (0 \cdot -3 - (-9) \cdot -3) - \hat{j} (4 \cdot -3 - (-9) \cdot 6) + \hat{k} (4 \cdot -3 - 0 \cdot 6) \]
\[ = \hat{i} (0 - 27) - \hat{j} (-12 + 54) + \hat{k} (-12) \]
\[ = -27\hat{i} - 42\hat{j} - 12\hat{k} \]

2. **Now calculate \( \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \)**:
\[ \vec{AB} \cdot (\vec{AC} \times \vec{AD}) = (-2, -2, -3) \cdot (-27, -42, -12) \]
\[ = (-2)(-27) + (-2)(-42) + (-3)(-12) \]
\[ = 54 + 84 + 36 = 174 \]

### Step 3: Calculate the volume of the tetrahedron

Using the scalar triple product in the volume formula:
\[ V = \frac{1}{6} |174| = \frac{174}{6} = 29 \]

Thus, the volume of the tetrahedron is \( \boxed{29} \) cubic units.