Find the value of `lambda` for which the four points with position vectors `6hati-7hatj, 16hati-19hatj, lambdahatj-6hatk and 2hati-5hatj+10hatk` are coplanar.

Answer

To find the value of \( \lambda \) for which the four points with position vectors \( \vec{A} = 6\hat{i} - 7\hat{j} \), \( \vec{B} = 16\hat{i} - 19\hat{j} \), \( \vec{C} = \lambda\hat{j} - 6\hat{k} \), and \( \vec{D} = 2\hat{i} - 5\hat{j} + 10\hat{k} \) are coplanar, we can use the condition that the vectors \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \) must be coplanar. This can be determined by checking if the scalar triple product of these vectors is zero.

### Step-by-Step Solution:

1. **Find the vectors \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \)**:
- \( \vec{AB} = \vec{B} - \vec{A} = (16\hat{i} - 19\hat{j}) - (6\hat{i} - 7\hat{j}) = (16 - 6)\hat{i} + (-19 + 7)\hat{j} = 10\hat{i} - 12\hat{j} \)
- \( \vec{AC} = \vec{C} - \vec{A} = (\lambda\hat{j} - 6\hat{k}) - (6\hat{i} - 7\hat{j}) = -6\hat{i} + (\lambda + 7)\hat{j} - 6\hat{k} \)
- \( \vec{AD} = \vec{D} - \vec{A} = (2\hat{i} - 5\hat{j} + 10\hat{k}) - (6\hat{i} - 7\hat{j}) = (2 - 6)\hat{i} + (-5 + 7)\hat{j} + 10\hat{k} = -4\hat{i} + 2\hat{j} + 10\hat{k} \)

2. **Set up the determinant for coplanarity**:
The points are coplanar if the determinant of the matrix formed by these vectors is zero:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 10 & -12 & 0 \\ -6 & \lambda + 7 & -6 \\ -4 & 2 & 10 \end{vmatrix} = 0 \]

3. **Calculate the determinant**:
Expanding the determinant:
\[ = \hat{i} \begin{vmatrix} -12 & 0 \\ \lambda + 7 & -6 \end{vmatrix} - \hat{j} \begin{vmatrix} 10 & 0 \\ -6 & -6 \end{vmatrix} + \hat{k} \begin{vmatrix} 10 & -12 \\ -6 & \lambda + 7 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
- For \( \hat{i} \):
\[ = -12 \cdot (-6) - 0 \cdot (\lambda + 7) = 72 \]
- For \( \hat{j} \):
\[ = 10 \cdot (-6) - 0 \cdot (-6) = -60 \]
- For \( \hat{k} \):
\[ = 10(\lambda + 7) - (-12)(-6) = 10\lambda + 70 - 72 = 10\lambda - 2 \]

Putting it all together:
\[ 72 + 60 + (10\lambda - 2) = 0 \]
\[ 130 + 10\lambda - 2 = 0 \]
\[ 10\lambda + 128 = 0 \]
\[ 10\lambda = -128 \]
\[ \lambda = -12.8 \]

### Final Answer:
The value of \( \lambda \) for which the four points are coplanar is \( \lambda = -12.8 \).