If the vectors, `P=(a+1)i+aj+ak,q=ai+(a+1)j+ak` and `r=ai+aj+(a+1)k,(a in R)` are coplanar and `3(p*q)^(2)-lamda|rxxq|^(2)=0` , then the value of `lamda` is _______ .

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the vectors
Given the vectors:
- \( \mathbf{p} = (a+1)\mathbf{i} + a\mathbf{j} + a\mathbf{k} \)
- \( \mathbf{q} = a\mathbf{i} + (a+1)\mathbf{j} + a\mathbf{k} \)
- \( \mathbf{r} = a\mathbf{i} + a\mathbf{j} + (a+1)\mathbf{k} \)

### Step 2: Check for coplanarity
Vectors \( \mathbf{p}, \mathbf{q}, \mathbf{r} \) are coplanar if the determinant of the matrix formed by their coefficients is zero:

\[ \begin{vmatrix} a+1 & a & a \\ a & a+1 & a \\ a & a & a+1 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
Calculating the determinant:

\[ D = (a+1) \begin{vmatrix} a+1 & a \\ a & a+1 \end{vmatrix} - a \begin{vmatrix} a & a \\ a & a+1 \end{vmatrix} + a \begin{vmatrix} a & a+1 \\ a & a \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} a+1 & a \\ a & a+1 \end{vmatrix} = (a+1)(a+1) - a^2 = a^2 + 2a + 1 - a^2 = 2a + 1 \)
2. \( \begin{vmatrix} a & a \\ a & a+1 \end{vmatrix} = a(a+1) - a^2 = a \)
3. \( \begin{vmatrix} a & a+1 \\ a & a \end{vmatrix} = a^2 - a(a+1) = -a \)

Substituting back into \( D \):

\[ D = (a+1)(2a + 1) - a(a) - a(-a) = (a+1)(2a + 1) - a^2 + a^2 = (a+1)(2a + 1) \]

Setting \( D = 0 \):

\[ (a+1)(2a + 1) = 0 \]

This gives us two cases:
1. \( a + 1 = 0 \) → \( a = -1 \)
2. \( 2a + 1 = 0 \) → \( a = -\frac{1}{2} \)

### Step 4: Calculate \( \mathbf{p} \cdot \mathbf{q} \) and \( \mathbf{r} \times \mathbf{q} \)
Now we need to find \( 3(\mathbf{p} \cdot \mathbf{q})^2 - \lambda |\mathbf{r} \times \mathbf{q}|^2 = 0 \).

#### Calculate \( \mathbf{p} \cdot \mathbf{q} \):
\[ \mathbf{p} \cdot \mathbf{q} = (a+1)a + a(a+1) + a^2 = a^2 + a + a^2 + a + a^2 = 3a^2 + 2a \]

#### Calculate \( |\mathbf{r} \times \mathbf{q}| \):
Using the determinant method for the cross product:

\[ \mathbf{r} \times \mathbf{q} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a & a & a + 1 \\ a & a + 1 & a \end{vmatrix} \]

Calculating this determinant gives us the components of \( \mathbf{r} \times \mathbf{q} \).

### Step 5: Substitute values into the equation
Substituting the values of \( \mathbf{p} \cdot \mathbf{q} \) and \( |\mathbf{r} \times \mathbf{q}| \) into the equation:

\[ 3(3a^2 + 2a)^2 - \lambda |\mathbf{r} \times \mathbf{q}|^2 = 0 \]

### Step 6: Solve for \( \lambda \)
From the equation, we can isolate \( \lambda \):

\[ \lambda = \frac{3(3a^2 + 2a)^2}{|\mathbf{r} \times \mathbf{q}|^2} \]

### Final Step: Find the value of \( \lambda \)
After calculating the specific values for \( a = -1 \) and \( a = -\frac{1}{2} \), we will find the corresponding values of \( \lambda \).