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 \( x - y + z + 1 = 0 \), \( \lambda x + 3y + 2z - 3 = 0 \), and \( 3x + \lambda y + z - 2 = 0 \) form a triangular prism, we need to ensure that the intersection lines of these planes are not parallel and that they form a closed triangular shape.

### Step 1: Identify the normal vectors of the planes
The normal vector of a plane given by the equation \( ax + by + cz + d = 0 \) is represented by the vector \( (a, b, c) \).

- For the first plane \( x - y + z + 1 = 0 \), the normal vector is \( \mathbf{n_1} = (1, -1, 1) \).
- For the second plane \( \lambda x + 3y + 2z - 3 = 0 \), the normal vector is \( \mathbf{n_2} = (\lambda, 3, 2) \).
- For the third plane \( 3x + \lambda y + z - 2 = 0 \), the normal vector is \( \mathbf{n_3} = (3, \lambda, 1) \).

### Step 2: Check the condition for non-parallel planes
For the planes to form a triangular prism, the normal vectors must not be coplanar. This can be checked using the scalar triple product:

\[ \mathbf{n_1} \cdot (\mathbf{n_2} \times \mathbf{n_3}) \neq 0 \]

### Step 3: Calculate the cross product \( \mathbf{n_2} \times \mathbf{n_3} \)
Using the determinant to find the cross product:

\[ \mathbf{n_2} \times \mathbf{n_3} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \lambda & 3 & 2 \\ 3 & \lambda & 1 \end{vmatrix} \]

Calculating the determinant:

\[ = \mathbf{i} \begin{vmatrix} 3 & 2 \\ \lambda & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} \lambda & 2 \\ 3 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} \lambda & 3 \\ 3 & \lambda \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( 3 \cdot 1 - 2 \cdot \lambda = 3 - 2\lambda \)
2. \( \lambda \cdot 1 - 2 \cdot 3 = \lambda - 6 \)
3. \( \lambda^2 - 9 \)

Thus, we have:

\[ \mathbf{n_2} \times \mathbf{n_3} = (3 - 2\lambda, -(\lambda - 6), \lambda^2 - 9) \]

### Step 4: Calculate the dot product \( \mathbf{n_1} \cdot (\mathbf{n_2} \times \mathbf{n_3}) \)
Now we compute:

\[ \mathbf{n_1} \cdot (\mathbf{n_2} \times \mathbf{n_3}) = (1, -1, 1) \cdot (3 - 2\lambda, -(\lambda - 6), \lambda^2 - 9) \]

Calculating this gives:

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

### Step 5: Set the expression to be non-zero
For the planes to not be coplanar, we need:

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

### Step 6: Solve the quadratic equation
To find the values of \( \lambda \) that make this expression zero, we solve:

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

Using the quadratic formula:

\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 + 48}}{2} = \frac{1 \pm 7}{2} \]

This gives:

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

### Step 7: Conclusion
Thus, the values of \( \lambda \) that make the planes coplanar are \( \lambda = 4 \) and \( \lambda = -3 \). Therefore, for the planes to form a triangular prism, \( \lambda \) must not equal these values.

### Final Answer
The value of \( \lambda \) should be any real number except \( 4 \) and \( -3 \).

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