The volume of the tetrahedron whose vertices are the points with position vectors `hat(i)+hat(j)+hat(k), -hat(i)-3hat(j)+7hat(k), hat(i)+2hat(j)-7hat(k) and 3hat(i)-4hat(j)+lambdahat(k)` is 22, then the digit at unit place of `lambda` is.

Answer

To find the value of \(\lambda\) such that the volume of the tetrahedron formed by the given vertices is 22, we can follow these steps:

### Step 1: Identify the position vectors
Let the vertices of the tetrahedron be represented by their position vectors:
- \( \mathbf{A} = \hat{i} + \hat{j} + \hat{k} \)
- \( \mathbf{B} = -\hat{i} - 3\hat{j} + 7\hat{k} \)
- \( \mathbf{C} = \hat{i} + 2\hat{j} - 7\hat{k} \)
- \( \mathbf{D} = 3\hat{i} - 4\hat{j} + \lambda\hat{k} \)

### Step 2: Calculate the position vectors relative to point A
We need to find the vectors \(\mathbf{AB}\), \(\mathbf{AC}\), and \(\mathbf{AD}\):
- \( \mathbf{AB} = \mathbf{B} - \mathbf{A} = (-\hat{i} - 3\hat{j} + 7\hat{k}) - (\hat{i} + \hat{j} + \hat{k}) = -2\hat{i} - 4\hat{j} + 6\hat{k} \)
- \( \mathbf{AC} = \mathbf{C} - \mathbf{A} = (\hat{i} + 2\hat{j} - 7\hat{k}) - (\hat{i} + \hat{j} + \hat{k}) = 0\hat{i} + \hat{j} - 8\hat{k} \)
- \( \mathbf{AD} = \mathbf{D} - \mathbf{A} = (3\hat{i} - 4\hat{j} + \lambda\hat{k}) - (\hat{i} + \hat{j} + \hat{k}) = 2\hat{i} - 5\hat{j} + (\lambda - 1)\hat{k} \)

### Step 3: Set up the volume formula
The volume \(V\) of the tetrahedron can be calculated using the formula:
\[ V = \frac{1}{6} \left| \mathbf{AB} \cdot (\mathbf{AC} \times \mathbf{AD}) \right| \]
We need to compute the cross product \(\mathbf{AC} \times \mathbf{AD}\).

### Step 4: Calculate the cross product \(\mathbf{AC} \times \mathbf{AD}\)
Using the vectors:
\[ \mathbf{AC} = \begin{pmatrix} 0 \\ 1 \\ -8 \end{pmatrix}, \quad \mathbf{AD} = \begin{pmatrix} 2 \\ -5 \\ \lambda - 1 \end{pmatrix} \]
The cross product is given by the determinant:
\[ \mathbf{AC} \times \mathbf{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & -8 \\ 2 & -5 & \lambda - 1 \end{vmatrix} \]
Calculating this determinant, we have:
\[ = \hat{i} \begin{vmatrix} 1 & -8 \\ -5 & \lambda - 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & -8 \\ 2 & \lambda - 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 1 \\ 2 & -5 \end{vmatrix} \]

Calculating each of these:
1. For \(\hat{i}\):
\[ 1(\lambda - 1) - (-8)(-5) = \lambda - 1 - 40 = \lambda - 41 \]
2. For \(-\hat{j}\):
\[ 0(\lambda - 1) - (-8)(2) = 16 \]
3. For \(\hat{k}\):
\[ 0(-5) - 1(2) = -2 \]

Thus,
\[ \mathbf{AC} \times \mathbf{AD} = (\lambda - 41)\hat{i} - 16\hat{j} - 2\hat{k} \]

### Step 5: Calculate the dot product \(\mathbf{AB} \cdot (\mathbf{AC} \times \mathbf{AD})\)
Now, we compute:
\[ \mathbf{AB} \cdot (\mathbf{AC} \times \mathbf{AD}) = \begin{pmatrix} -2 \\ -4 \\ 6 \end{pmatrix} \cdot \begin{pmatrix} \lambda - 41 \\ -16 \\ -2 \end{pmatrix} \]
Calculating this gives:
\[ = -2(\lambda - 41) - 4(-16) + 6(-2) \]
\[ = -2\lambda + 82 + 64 - 12 = -2\lambda + 134 \]

### Step 6: Set the volume equal to 22
Setting the volume equal to 22:
\[ \frac{1}{6} |-2\lambda + 134| = 22 \]
Multiplying both sides by 6:
\[ |-2\lambda + 134| = 132 \]
This gives two equations:
1. \( -2\lambda + 134 = 132 \)
2. \( -2\lambda + 134 = -132 \)

### Step 7: Solve for \(\lambda\)
1. From \( -2\lambda + 134 = 132 \):
\[ -2\lambda = -2 \implies \lambda = 1 \]
2. From \( -2\lambda + 134 = -132 \):
\[ -2\lambda = -266 \implies \lambda = 133 \]

### Step 8: Find the unit place of \(\lambda\)
The possible values for \(\lambda\) are 1 and 133. The unit place of \(\lambda\) is:
- For \(\lambda = 1\), the unit place is 1.
- For \(\lambda = 133\), the unit place is 3.

Since the problem asks for the unit place of \(\lambda\), we can conclude that the answer is **1**.