If the volume of the tetrahedron whose vertices are `(1,-6,10),(-1,-3,7),(5,-1,lamda)` and `(7,-4,7)` is 11 cubit units then `lamda=`

Options

A. 2,6
B. 3,4
C. 1,7
D. 5,6

Explanation

To find the value of \( \lambda \) for the tetrahedron with vertices \( A(1, -6, 10) \), \( B(-1, -3, 7) \), \( C(5, -1, \lambda) \), and \( D(7, -4, 7) \) such that the volume is 11 cubic units, we can follow these steps:

### Step 1: Determine the vectors
First, we need to find the vectors corresponding to the edges of the tetrahedron that share a common vertex. We can choose vertex \( A \) as the common vertex.

- Vector \( \vec{AB} = B - A = (-1 - 1, -3 + 6, 7 - 10) = (-2, 3, -3) \)
- Vector \( \vec{AC} = C - A = (5 - 1, -1 + 6, \lambda - 10) = (4, 5, \lambda - 10) \)
- Vector \( \vec{AD} = D - A = (7 - 1, -4 + 6, 7 - 10) = (6, 2, -3) \)

### Step 2: Set up the volume formula
The volume \( V \) of a tetrahedron formed by vectors \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \) is given by:
\[ V = \frac{1}{6} | \vec{AB} \cdot (\vec{AC} \times \vec{AD}) | \]

### Step 3: Calculate the cross product \( \vec{AC} \times \vec{AD} \)
To find the cross product \( \vec{AC} \times \vec{AD} \):
\[ \vec{AC} = (4, 5, \lambda - 10), \quad \vec{AD} = (6, 2, -3) \]
Using the determinant formula for the cross product:
\[ \vec{AC} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & \lambda - 10 \\ 6 & 2 & -3 \end{vmatrix} \]

Calculating the 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 each of the 2x2 determinants:
1. \( \begin{vmatrix} 5 & \lambda - 10 \\ 2 & -3 \end{vmatrix} = 5(-3) - 2(\lambda - 10) = -15 - 2\lambda + 20 = 5 - 2\lambda \)
2. \( \begin{vmatrix} 4 & \lambda - 10 \\ 6 & -3 \end{vmatrix} = 4(-3) - 6(\lambda - 10) = -12 - 6\lambda + 60 = 48 - 6\lambda \)
3. \( \begin{vmatrix} 4 & 5 \\ 6 & 2 \end{vmatrix} = 4(2) - 5(6) = 8 - 30 = -22 \)

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

### Step 4: Calculate the dot product \( \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \)
Now, we compute:
\[ \vec{AB} \cdot (\vec{AC} \times \vec{AD}) = (-2, 3, -3) \cdot (5 - 2\lambda, -(48 - 6\lambda), -22) \]
Calculating the dot product:
\[ = -2(5 - 2\lambda) + 3(-(48 - 6\lambda)) - 3(-22) \]
\[ = -10 + 4\lambda - 144 + 18\lambda + 66 \]
\[ = 22\lambda - 88 \]

### Step 5: Set up the volume equation
The volume is given as 11 cubic units:
\[ \frac{1}{6} |22\lambda - 88| = 11 \]
Multiplying both sides by 6:
\[ |22\lambda - 88| = 66 \]

### Step 6: Solve the absolute value equation
This gives us two cases to solve:
1. \( 22\lambda - 88 = 66 \)
2. \( 22\lambda - 88 = -66 \)

**For the first case:**
\[ 22\lambda = 154 \implies \lambda = \frac{154}{22} = 7 \]

**For the second case:**
\[ 22\lambda = 22 \implies \lambda = 1 \]

### Final Answer
Thus, the values of \( \lambda \) are \( \lambda = 1 \) and \( \lambda = 7 \).