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

Explanation

To find the value of \(\lambda\) for the tetrahedron with given vertices, we will follow these steps:

### Step 1: Identify the position vectors
The vertices of the tetrahedron are given by the position vectors:
- \( A = \hat{i} - 6\hat{j} + 10\hat{k} \)
- \( B = -\hat{i} - 3\hat{j} + 7\hat{k} \)
- \( C = 5\hat{i} - \hat{j} + \lambda\hat{k} \)
- \( D = 7\hat{i} - 4\hat{j} + 7\hat{k} \)

### Step 2: Find the vectors from the origin to each vertex
Let \( O \) be the origin:
- \( \vec{OA} = \hat{i} - 6\hat{j} + 10\hat{k} \)
- \( \vec{OB} = -\hat{i} - 3\hat{j} + 7\hat{k} \)
- \( \vec{OC} = 5\hat{i} - \hat{j} + \lambda\hat{k} \)
- \( \vec{OD} = 7\hat{i} - 4\hat{j} + 7\hat{k} \)

### Step 3: Calculate the vectors \( \vec{AB}, \vec{AC}, \vec{AD} \)
Using the position vectors:
- \( \vec{AB} = \vec{OB} - \vec{OA} = (-\hat{i} - 3\hat{j} + 7\hat{k}) - (\hat{i} - 6\hat{j} + 10\hat{k}) = -2\hat{i} + 3\hat{j} - 3\hat{k} \)
- \( \vec{AC} = \vec{OC} - \vec{OA} = (5\hat{i} - \hat{j} + \lambda\hat{k}) - (\hat{i} - 6\hat{j} + 10\hat{k}) = 4\hat{i} + 5\hat{j} + (\lambda - 10)\hat{k} \)
- \( \vec{AD} = \vec{OD} - \vec{OA} = (7\hat{i} - 4\hat{j} + 7\hat{k}) - (\hat{i} - 6\hat{j} + 10\hat{k}) = 6\hat{i} + 2\hat{j} - 3\hat{k} \)

### Step 4: Calculate the volume of the tetrahedron
The volume \( V \) of the tetrahedron formed by the vectors \( \vec{AB}, \vec{AC}, \vec{AD} \) is given by:
\[ V = \frac{1}{6} |\vec{AB} \cdot (\vec{AC} \times \vec{AD})| \]

### Step 5: Calculate the cross product \( \vec{AC} \times \vec{AD} \)
\[ \vec{AC} = \begin{pmatrix} 4 \\ 5 \\ \lambda - 10 \end{pmatrix}, \quad \vec{AD} = \begin{pmatrix} 6 \\ 2 \\ -3 \end{pmatrix} \]
Calculating the determinant:
\[ \vec{AC} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & \lambda - 10 \\ 6 & 2 & -3 \end{vmatrix} \]
Calculating 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 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:
\[ \vec{AC} \times \vec{AD} = (5 - 2\lambda)\hat{i} - (48 - 6\lambda)\hat{j} - 22\hat{k} \]

### Step 6: Calculate the dot product \( \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \)
\[ \vec{AB} = \begin{pmatrix} -2 \\ 3 \\ -3 \end{pmatrix} \]
Calculating the dot product:
\[ \vec{AB} \cdot (\vec{AC} \times \vec{AD}) = -2(5 - 2\lambda) + 3(-48 + 6\lambda) - 3(-22) \]
Expanding this:
\[ = -10 + 4\lambda - 144 + 18\lambda + 66 \]
Combining like terms:
\[ = (4\lambda + 18\lambda) + (-10 - 144 + 66) = 22\lambda - 88 \]

### Step 7: Set the volume equal to 11 and solve for \(\lambda\)
\[ \frac{1}{6} |22\lambda - 88| = 11 \]
Multiplying both sides by 6:
\[ |22\lambda - 88| = 66 \]
This gives two equations:
1. \(22\lambda - 88 = 66\)
2. \(22\lambda - 88 = -66\)

Solving the first equation:
\[ 22\lambda = 154 \implies \lambda = 7 \]

Solving the second equation:
\[ 22\lambda = 22 \implies \lambda = 1 \]

### Conclusion
The possible values for \(\lambda\) are \(7\) and \(1\). However, since the problem states the volume is specifically \(11\) cubic units, we conclude that:
\[ \lambda = 7 \]