The volume of the tetrahedron whose vertices are points `A(1,-1,10), B (-1,-3,7), C(5, -1,lambda), D(7,-4,7)` be 11 cubic units then the value of `lambda` is

Options

A. `-1`
B. 1
C. `-7`
D. 7

Answer

To find the value of \( \lambda \) for the tetrahedron with vertices \( A(1, -1, 10) \), \( B(-1, -3, 7) \), \( C(5, -1, \lambda) \), and \( D(7, -4, 7) \) such that its volume is 11 cubic units, 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: Calculate 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 \( \vec{AC} \)**:
\[ \vec{AC} = C - A = (5 - 1, -1 - (-1), \lambda - 10) = (4, 0, \lambda - 10) \]

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

### Step 2: Set up the determinant for the scalar triple product

The volume is given by:
\[ V = \frac{1}{6} | \vec{AB} \cdot (\vec{AC} \times \vec{AD}) | \]

We need to compute the scalar triple product \( \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \). This can be calculated using the determinant:

\[ \begin{vmatrix} -2 & -2 & -3 \\ 4 & 0 & \lambda - 10 \\ 6 & -3 & -3 \end{vmatrix} \]

### Step 3: Calculate the determinant

Using the determinant formula:
\[ \text{Det} = -2 \begin{vmatrix} 0 & \lambda - 10 \\ -3 & -3 \end{vmatrix} + 2 \begin{vmatrix} 4 & \lambda - 10 \\ 6 & -3 \end{vmatrix} - 3 \begin{vmatrix} 4 & 0 \\ 6 & -3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. **First determinant**:
\[ \begin{vmatrix} 0 & \lambda - 10 \\ -3 & -3 \end{vmatrix} = 0 \cdot (-3) - (-3)(\lambda - 10) = 3(\lambda - 10) \]

2. **Second determinant**:
\[ \begin{vmatrix} 4 & \lambda - 10 \\ 6 & -3 \end{vmatrix} = 4(-3) - 6(\lambda - 10) = -12 - 6\lambda + 60 = 48 - 6\lambda \]

3. **Third determinant**:
\[ \begin{vmatrix} 4 & 0 \\ 6 & -3 \end{vmatrix} = 4(-3) - 0 \cdot 6 = -12 \]

Putting it all together:
\[ \text{Det} = -2(3(\lambda - 10)) + 2(48 - 6\lambda) - 3(-12) \]
\[ = -6(\lambda - 10) + 96 - 12\lambda + 36 \]
\[ = -6\lambda + 60 + 96 - 12\lambda + 36 \]
\[ = -18\lambda + 192 \]

### Step 4: Set the volume equal to 11 cubic units

Since the volume is given as 11 cubic units:
\[ \frac{1}{6} |-18\lambda + 192| = 11 \]
Multiplying both sides by 6:
\[ |-18\lambda + 192| = 66 \]

### Step 5: Solve for \( \lambda \)

This gives us two equations:
1. \( -18\lambda + 192 = 66 \)
2. \( -18\lambda + 192 = -66 \)

**For the first equation**:
\[ -18\lambda = 66 - 192 \]
\[ -18\lambda = -126 \implies \lambda = \frac{126}{18} = 7 \]

**For the second equation**:
\[ -18\lambda = -66 - 192 \]
\[ -18\lambda = -258 \implies \lambda = \frac{258}{18} = \frac{43}{3} \]

### Conclusion

The possible values for \( \lambda \) are \( 7 \) and \( \frac{43}{3} \).