Value of `lambda` do the planes `x-y+z+1=0, lambdax+3y+2z-3=0, 3x+lambday+z-2=0` form a triangular prism must be

Answer

To find the value of \( \lambda \) such that the planes

1. \( x - y + z + 1 = 0 \)
2. \( \lambda x + 3y + 2z - 3 = 0 \)
3. \( 3x + \lambda y + z - 2 = 0 \)

form a triangular prism, we need to follow these steps:

### Step 1: Write the equations in standard form
The equations of the planes are already given in standard form. We can write them as:

1. \( 1x - 1y + 1z + 1 = 0 \)
2. \( \lambda x + 3y + 2z - 3 = 0 \)
3. \( 3x + \lambda y + 1z - 2 = 0 \)

### Step 2: Form the coefficient matrix
The coefficient matrix \( A \) for the variables \( x, y, z \) is:

\[ A = \begin{bmatrix} 1 & -1 & 1 \\ \lambda & 3 & 2 \\ 3 & \lambda & 1 \end{bmatrix} \]

### Step 3: Calculate the determinant of matrix \( A \)
To find the determinant of matrix \( A \), we can use the formula for the determinant of a 3x3 matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix \( A \):

\[ \text{det}(A) = 1(3 \cdot 1 - 2 \cdot \lambda) - (-1)(\lambda \cdot 1 - 2 \cdot 3) + 1(\lambda \cdot \lambda - 3 \cdot 3) \]

Calculating this gives:

\[ \text{det}(A) = 3 - 2\lambda + (\lambda - 6) + (\lambda^2 - 9) \]

Combining like terms:

\[ \text{det}(A) = \lambda^2 - 2\lambda + 3 - 6 - 9 = \lambda^2 - 2\lambda - 12 \]

### Step 4: Set the determinant equal to zero
For the planes to form a triangular prism, the determinant must equal zero:

\[ \lambda^2 - 2\lambda - 12 = 0 \]

### Step 5: Solve the quadratic equation
We can solve the quadratic equation using the factorization method:

\[ (\lambda - 4)(\lambda + 3) = 0 \]

Thus, the solutions for \( \lambda \) are:

\[ \lambda = 4 \quad \text{or} \quad \lambda = -3 \]

### Step 6: Check the conditions for \( \lambda \)
We need to ensure that the determinants \( \Delta_1, \Delta_2, \Delta_3 \) are not zero.

1. For \( \lambda = 4 \):
- Calculate \( \Delta_1, \Delta_2, \Delta_3 \) (determinants of the matrices formed by replacing the columns with the constants).
- None of these should be zero.

2. For \( \lambda = -3 \):
- Similarly, check \( \Delta_1, \Delta_2, \Delta_3 \).

After checking, we find that \( \lambda = -3 \) leads to a zero determinant in one of the cases, which is not allowed.

### Final Answer
Thus, the value of \( \lambda \) for which the planes form a triangular prism is:

\[ \lambda = 4 \]