If `vec P = (a+1) hati + a hatj + a hatk` ` vec Q = a hati + (a+1) hatj + ahatk` `vec R = a hati + a hatj + (a+1) hatk` ` vec P, vec Q, vec R` are coplanar vectors and `3(vec P . vec Q)^2 - lambda | vec R xx vec Q|^2 = 0` then value of `lambda` is

Explanation

To solve the problem, we need to find the value of \( \lambda \) given the vectors \( \vec{P} \), \( \vec{Q} \), and \( \vec{R} \) are coplanar and satisfy the equation:

\[ 3(\vec{P} \cdot \vec{Q})^2 - \lambda |\vec{R} \times \vec{Q}|^2 = 0 \]

### Step 1: Define the vectors

The vectors are given as:

\[ \vec{P} = (a+1) \hat{i} + a \hat{j} + a \hat{k} \]
\[ \vec{Q} = a \hat{i} + (a+1) \hat{j} + a \hat{k} \]
\[ \vec{R} = a \hat{i} + a \hat{j} + (a+1) \hat{k} \]

### Step 2: Check for coplanarity

For the vectors \( \vec{P} \), \( \vec{Q} \), and \( \vec{R} \) to be coplanar, the scalar triple product must be zero:

\[ \vec{P} \cdot (\vec{Q} \times \vec{R}) = 0 \]

### Step 3: Calculate \( \vec{Q} \times \vec{R} \)

To find \( \vec{Q} \times \vec{R} \), we set up the determinant:

\[ \vec{Q} \times \vec{R} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a & a+1 & a \\ a & a & a+1 \end{vmatrix} \]

Calculating this determinant:

\[ = \hat{i} \left( (a+1)(a+1) - a^2 \right) - \hat{j} \left( a(a+1) - a^2 \right) + \hat{k} \left( a(a) - a(a+1) \right) \]

This simplifies to:

\[ = \hat{i} \left( a^2 + 2a + 1 - a^2 \right) - \hat{j} \left( a^2 + a - a^2 \right) + \hat{k} \left( a^2 - a^2 - a \right) \]

\[ = (2a + 1) \hat{i} - a \hat{j} - a \hat{k} \]

### Step 4: Calculate \( |\vec{Q} \times \vec{R}|^2 \)

The magnitude squared is:

\[ |\vec{Q} \times \vec{R}|^2 = (2a + 1)^2 + (-a)^2 + (-a)^2 = (2a + 1)^2 + 2a^2 \]

Expanding this:

\[ = 4a^2 + 4a + 1 + 2a^2 = 6a^2 + 4a + 1 \]

### Step 5: Calculate \( \vec{P} \cdot \vec{Q} \)

Now, we calculate the dot product \( \vec{P} \cdot \vec{Q} \):

\[ \vec{P} \cdot \vec{Q} = ((a+1)a) + (a(a+1)) + (a^2) \]

This simplifies to:

\[ = a^2 + a + a^2 + a + a^2 = 3a^2 + 2a \]

### Step 6: Substitute into the equation

Now substituting into the equation:

\[ 3(\vec{P} \cdot \vec{Q})^2 - \lambda |\vec{Q} \times \vec{R}|^2 = 0 \]

This becomes:

\[ 3(3a^2 + 2a)^2 - \lambda (6a^2 + 4a + 1) = 0 \]

### Step 7: Solve for \( \lambda \)

Expanding \( (3a^2 + 2a)^2 \):

\[ = 9a^4 + 12a^3 + 4a^2 \]

Thus:

\[ 3(9a^4 + 12a^3 + 4a^2) - \lambda (6a^2 + 4a + 1) = 0 \]

This simplifies to:

\[ 27a^4 + 36a^3 + 12a^2 - \lambda (6a^2 + 4a + 1) = 0 \]

### Step 8: Find \( \lambda \)

For this to hold for all \( a \), we can equate coefficients. The coefficient of \( a^2 \) gives:

\[ 12 - 6\lambda = 0 \implies \lambda = 2 \]

Thus, the value of \( \lambda \) is:

\[ \lambda = 2 \]

### Final Answer:

\[ \lambda = 2 \]