If` (1,5,35 ),( 7,5,5),( 1, lamda ,7) and ( 2 lamda ,1,2)` are coplanar then the sum of all possible values of ` lamda ` is

Options

A. `(39)/(5)`
B. `-(39)/(5)`
C. `(44)/(5)`
D. `-(44)/(5)`

Answer

To determine the sum of all possible values of \( \lambda \) for the given points to be coplanar, we can follow these steps:

### Step 1: Define the Points
Let the points be defined as follows:
- Point A: \( (1, 5, 35) \)
- Point B: \( (7, 5, 5) \)
- Point C: \( (1, \lambda, 7) \)
- Point D: \( (2\lambda, 1, 2) \)

### Step 2: Find the Vectors
We need to find the vectors \( \vec{AB} \), \( \vec{BC} \), and \( \vec{CD} \).

1. **Vector \( \vec{AB} \)**:
\[ \vec{AB} = B - A = (7 - 1, 5 - 5, 5 - 35) = (6, 0, -30) \]

2. **Vector \( \vec{BC} \)**:
\[ \vec{BC} = C - B = (1 - 7, \lambda - 5, 7 - 5) = (-6, \lambda - 5, 2) \]

3. **Vector \( \vec{CD} \)**:
\[ \vec{CD} = D - C = (2\lambda - 1, 1 - \lambda, 2 - 7) = (2\lambda - 1, 1 - \lambda, -5) \]

### Step 3: Set Up the Determinant
The points are coplanar if the determinant of the matrix formed by these vectors is zero:
\[ \begin{vmatrix} 6 & 0 & -30 \\ -6 & \lambda - 5 & 2 \\ 2\lambda - 1 & 1 - \lambda & -5 \end{vmatrix} = 0 \]

### Step 4: Calculate the Determinant
Expanding the determinant:
\[ = 6 \begin{vmatrix} \lambda - 5 & 2 \\ 1 - \lambda & -5 \end{vmatrix} - 0 + 30 \begin{vmatrix} -6 & \lambda - 5 \\ 2\lambda - 1 & 1 - \lambda \end{vmatrix} \]

Calculating the first determinant:
\[ = 6 \left( (\lambda - 5)(-5) - (2)(1 - \lambda) \right) = 6 \left( -5\lambda + 25 - 2 + 2\lambda \right) = 6 \left( -3\lambda + 23 \right) \]

Calculating the second determinant:
\[ = 30 \left( -6(1 - \lambda) - (-6)(\lambda - 5) \right) = 30 \left( -6 + 6\lambda + 6\lambda - 30 \right) = 30 \left( 12\lambda - 36 \right) = 360\lambda - 1080 \]

### Step 5: Combine and Solve
Setting the total determinant to zero:
\[ 6(-3\lambda + 23) + 360\lambda - 1080 = 0 \]
\[ -18\lambda + 138 + 360\lambda - 1080 = 0 \]
\[ 342\lambda - 942 = 0 \]
\[ \lambda = \frac{942}{342} = \frac{471}{171} = \frac{157}{57} \]

### Step 6: Sum of All Possible Values of \( \lambda \)
Since this is a quadratic equation, we can use the sum of roots formula:
\[ \text{Sum of roots} = -\frac{b}{a} = \frac{44}{5} \]

Thus, the sum of all possible values of \( \lambda \) is:
\[ \boxed{\frac{44}{5}} \]