Let the position vectors of the points A, B, C and D be `5hati+5j+2lambda hatk, hati + 2hatj + 3hatk, -2hati +lambda hat j+4hat k` and `-hati +5hatj +6hatk` .Let the set `S={lambda in RR `: the points A, B , C and D are complanar}. The `sum_(lambda in S)(lambda + 2)^2` is equal to

Options

A. 25
B. `(37)/2`
C. 13
D. 41

Answer

To solve the problem, we need to determine the values of \( \lambda \) for which the points A, B, C, and D are coplanar. The position vectors of the points A, B, C, and D are given as:

- \( \vec{A} = 5\hat{i} + 5\hat{j} + 2\lambda\hat{k} \)
- \( \vec{B} = \hat{i} + 2\hat{j} + 3\hat{k} \)
- \( \vec{C} = -2\hat{i} + \lambda\hat{j} + 4\hat{k} \)
- \( \vec{D} = -\hat{i} + 5\hat{j} + 6\hat{k} \)

### Step 1: Find the vectors AB, AC, and AD
We first calculate the vectors \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \):

\[ \vec{AB} = \vec{B} - \vec{A} = (\hat{i} + 2\hat{j} + 3\hat{k}) - (5\hat{i} + 5\hat{j} + 2\lambda\hat{k}) = (-4\hat{i} - 3\hat{j} + (3 - 2\lambda)\hat{k}) \]

\[ \vec{AC} = \vec{C} - \vec{A} = (-2\hat{i} + \lambda\hat{j} + 4\hat{k}) - (5\hat{i} + 5\hat{j} + 2\lambda\hat{k}) = (-7\hat{i} + (\lambda - 5)\hat{j} + (4 - 2\lambda)\hat{k}) \]

\[ \vec{AD} = \vec{D} - \vec{A} = (-\hat{i} + 5\hat{j} + 6\hat{k}) - (5\hat{i} + 5\hat{j} + 2\lambda\hat{k}) = (-6\hat{i} + 0\hat{j} + (6 - 2\lambda)\hat{k}) \]

### Step 2: Set up the coplanarity condition
The points A, B, C, and D are coplanar if the scalar triple product \( \vec{AB} \cdot (\vec{AC} \times \vec{AD}) = 0 \).

### Step 3: Calculate the cross product \( \vec{AC} \times \vec{AD} \)
Using the determinant method to find the cross product:

\[ \vec{AC} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -7 & \lambda - 5 & 4 - 2\lambda \\ -6 & 0 & 6 - 2\lambda \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \left( (\lambda - 5)(6 - 2\lambda) - 0 \cdot (4 - 2\lambda) \right) - \hat{j} \left( -7(6 - 2\lambda) - (-6)(4 - 2\lambda) \right) + \hat{k} \left( -7 \cdot 0 - (-6)(\lambda - 5) \right) \]

\[ = \hat{i} \left( 6\lambda - 30 - 12\lambda + 10\lambda \right) - \hat{j} \left( -42 + 14\lambda + 24 - 12\lambda \right) + \hat{k} \left( 6\lambda - 30 \right) \]

\[ = \hat{i} \left( -6\lambda - 30 \right) - \hat{j} \left( 2\lambda - 18 \right) + \hat{k} \left( 6\lambda - 30 \right) \]

### Step 4: Dot product with \( \vec{AB} \)
Now, we compute \( \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \):

\[ \vec{AB} \cdot (\vec{AC} \times \vec{AD}) = (-4\hat{i} - 3\hat{j} + (3 - 2\lambda)\hat{k}) \cdot \left( -6\lambda - 30, -2\lambda + 18, 6\lambda - 30 \right) \]

Calculating this dot product gives us:

\[ = -4(-6\lambda - 30) - 3(-2\lambda + 18) + (3 - 2\lambda)(6\lambda - 30) \]

### Step 5: Set the equation to zero
Setting the above expression to zero will yield a quadratic equation in \( \lambda \):

\[ \text{(After simplifying)} \rightarrow 4\lambda^2 - 20\lambda + 36 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ \lambda = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 4 \cdot 36}}{2 \cdot 4} \]

Calculating the discriminant and solving will give us the values of \( \lambda \).

### Step 7: Calculate the required sum
Once we have the values of \( \lambda \), we compute \( \sum_{\lambda \in S} (\lambda + 2)^2 \).