The volume of the tetrahedron whose vertices are the points `hati, hati+hatj, hati+hatj+hatk` and `2hati+3hatj+lamdahatk` is `1//6` units,
Then the values of `lamda`

Options

A. does not exist
B. is 7
C. is -1
D. is any real value

Explanation

To find the value of \( \lambda \) such that the volume of the tetrahedron with vertices at the points \( \mathbf{a} = \hat{i} \), \( \mathbf{b} = \hat{i} + \hat{j} \), \( \mathbf{c} = \hat{i} + \hat{j} + \hat{k} \), and \( \mathbf{d} = 2\hat{i} + 3\hat{j} + \lambda \hat{k} \) is \( \frac{1}{6} \) units, we can follow these steps:

### Step 1: Identify the vectors
We define the vertices as:
- \( \mathbf{a} = \hat{i} \)
- \( \mathbf{b} = \hat{i} + \hat{j} \)
- \( \mathbf{c} = \hat{i} + \hat{j} + \hat{k} \)
- \( \mathbf{d} = 2\hat{i} + 3\hat{j} + \lambda \hat{k} \)

### Step 2: Calculate the vectors \( \mathbf{AB} \), \( \mathbf{AC} \), and \( \mathbf{AD} \)
- \( \mathbf{AB} = \mathbf{b} - \mathbf{a} = (\hat{i} + \hat{j}) - \hat{i} = \hat{j} \)
- \( \mathbf{AC} = \mathbf{c} - \mathbf{a} = (\hat{i} + \hat{j} + \hat{k}) - \hat{i} = \hat{j} + \hat{k} \)
- \( \mathbf{AD} = \mathbf{d} - \mathbf{a} = (2\hat{i} + 3\hat{j} + \lambda \hat{k}) - \hat{i} = \hat{i} + 3\hat{j} + \lambda \hat{k} \)

### Step 3: Set up the scalar triple product
The volume \( V \) of the tetrahedron can be expressed as:
\[ V = \frac{1}{6} |\mathbf{AB} \cdot (\mathbf{AC} \times \mathbf{AD})| \]
Given that the volume is \( \frac{1}{6} \), we have:
\[ |\mathbf{AB} \cdot (\mathbf{AC} \times \mathbf{AD})| = 1 \]

### Step 4: Compute the cross product \( \mathbf{AC} \times \mathbf{AD} \)
We can represent the vectors in component form:
- \( \mathbf{AB} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \)
- \( \mathbf{AC} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \)
- \( \mathbf{AD} = \begin{pmatrix} 1 \\ 3 \\ \lambda \end{pmatrix} \)

Now, we calculate the cross product \( \mathbf{AC} \times \mathbf{AD} \):
\[ \mathbf{AC} \times \mathbf{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 1 & 3 & \lambda \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ 3 & \lambda \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 1 \\ 1 & \lambda \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 1 \\ 1 & 3 \end{vmatrix} \]
Calculating each of these determinants:
- For \( \hat{i} \): \( 1 \cdot \lambda - 1 \cdot 3 = \lambda - 3 \)
- For \( \hat{j} \): \( 0 \cdot \lambda - 1 \cdot 1 = -1 \)
- For \( \hat{k} \): \( 0 \cdot 3 - 1 \cdot 1 = -1 \)

Thus, we have:
\[ \mathbf{AC} \times \mathbf{AD} = (\lambda - 3)\hat{i} + 1\hat{j} - 1\hat{k} \]

### Step 5: Calculate the dot product \( \mathbf{AB} \cdot (\mathbf{AC} \times \mathbf{AD}) \)
Now, we compute:
\[ \mathbf{AB} \cdot (\mathbf{AC} \times \mathbf{AD}) = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} \lambda - 3 \\ 1 \\ -1 \end{pmatrix} = 0 \cdot (\lambda - 3) + 1 \cdot 1 + 0 \cdot (-1) = 1 \]

### Step 6: Set the equation equal to 1
Since we have:
\[ |\mathbf{AB} \cdot (\mathbf{AC} \times \mathbf{AD})| = 1 \]
This is already satisfied, and we find that the equation holds true regardless of the value of \( \lambda \).

### Conclusion
Since the equation holds for any value of \( \lambda \), we conclude that:
\[ \lambda \in \mathbb{R} \]