The volume of the tetrahedron whose vertices are the points with positon vectors `hati-6hatj+10hatk, -hati-3hatj+7hatk, 5hati-hatj+lambdahatk` and `7hati-4hatj+7hatk` is 11 cubic units if the value of `lamda` is

Options

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

Explanation

To find the value of \( \lambda \) for the tetrahedron with given vertices, we will use the formula for the volume of a tetrahedron defined by vectors. The volume \( V \) of a tetrahedron formed by the position vectors \( \vec{A}, \vec{B}, \vec{C}, \vec{D} \) is given by:

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

Where:
- \( \vec{AB} = \vec{B} - \vec{A} \)
- \( \vec{AC} = \vec{C} - \vec{A} \)
- \( \vec{AD} = \vec{D} - \vec{A} \)

### Step 1: Define the position vectors
Let:
- \( \vec{A} = \hat{i} - 6\hat{j} + 10\hat{k} \)
- \( \vec{B} = -\hat{i} - 3\hat{j} + 7\hat{k} \)
- \( \vec{C} = 5\hat{i} - \hat{j} + \lambda\hat{k} \)
- \( \vec{D} = 7\hat{i} - 4\hat{j} + 7\hat{k} \)

### Step 2: Calculate the vectors \( \vec{AB}, \vec{AC}, \vec{AD} \)
1. **Calculate \( \vec{AB} \)**:
\[ \vec{AB} = \vec{B} - \vec{A} = (-\hat{i} - 3\hat{j} + 7\hat{k}) - (\hat{i} - 6\hat{j} + 10\hat{k}) = -2\hat{i} + 3\hat{j} - 3\hat{k} \]

2. **Calculate \( \vec{AC} \)**:
\[ \vec{AC} = \vec{C} - \vec{A} = (5\hat{i} - \hat{j} + \lambda\hat{k}) - (\hat{i} - 6\hat{j} + 10\hat{k}) = 4\hat{i} + 5\hat{j} + (\lambda - 10)\hat{k} \]

3. **Calculate \( \vec{AD} \)**:
\[ \vec{AD} = \vec{D} - \vec{A} = (7\hat{i} - 4\hat{j} + 7\hat{k}) - (\hat{i} - 6\hat{j} + 10\hat{k}) = 6\hat{i} + 2\hat{j} - 3\hat{k} \]

### Step 3: Calculate the scalar triple product \( \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \)
1. **Calculate \( \vec{AC} \times \vec{AD} \)**:
\[ \vec{AC} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & \lambda - 10 \\ 6 & 2 & -3 \end{vmatrix} \]

Expanding this determinant:
\[ = \hat{i} \begin{vmatrix} 5 & \lambda - 10 \\ 2 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & \lambda - 10 \\ 6 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & 5 \\ 6 & 2 \end{vmatrix} \]

Calculating the minors:
\[ = \hat{i} (5 \cdot (-3) - 2 \cdot (\lambda - 10)) - \hat{j} (4 \cdot (-3) - 6 \cdot (\lambda - 10)) + \hat{k} (4 \cdot 2 - 5 \cdot 6) \]
\[ = \hat{i} (-15 - 2\lambda + 20) - \hat{j} (-12 - 6\lambda + 60) + \hat{k} (8 - 30) \]
\[ = \hat{i} (5 - 2\lambda) - \hat{j} (48 - 6\lambda) - 22\hat{k} \]

Thus,
\[ \vec{AC} \times \vec{AD} = (5 - 2\lambda)\hat{i} - (48 - 6\lambda)\hat{j} - 22\hat{k} \]

2. **Calculate \( \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \)**:
\[ \vec{AB} \cdot (\vec{AC} \times \vec{AD}) = (-2\hat{i} + 3\hat{j} - 3\hat{k}) \cdot ((5 - 2\lambda)\hat{i} - (48 - 6\lambda)\hat{j} - 22\hat{k}) \]
\[ = -2(5 - 2\lambda) + 3(-48 + 6\lambda) - 3(-22) \]
\[ = -10 + 4\lambda - 144 + 18\lambda + 66 \]
\[ = 22\lambda - 88 \]

### Step 4: Set the volume equal to 11 cubic units
Using the volume formula:
\[ \frac{1}{6} |22\lambda - 88| = 11 \]
\[ |22\lambda - 88| = 66 \]

This gives us two equations:
1. \( 22\lambda - 88 = 66 \)
2. \( 22\lambda - 88 = -66 \)

### Step 5: Solve for \( \lambda \)
1. From \( 22\lambda - 88 = 66 \):
\[ 22\lambda = 154 \implies \lambda = \frac{154}{22} = 7 \]

2. From \( 22\lambda - 88 = -66 \):
\[ 22\lambda = 22 \implies \lambda = 1 \]

### Final Answer
The values of \( \lambda \) are \( 1 \) and \( 7 \).