If the system of equations `3x-2y+z=0, lambda x-14y+15z=0, x+2y+3z=0` have a non trivial solution, then the value of `lambda^(2)` must be

Explanation

To find the value of \( \lambda^2 \) such that the system of equations has a non-trivial solution, we need to analyze the given equations:

1. \( 3x - 2y + z = 0 \) (Equation 1)
2. \( \lambda x - 14y + 15z = 0 \) (Equation 2)
3. \( x + 2y + 3z = 0 \) (Equation 3)

### Step 1: Form the Coefficient Matrix
We can represent the system of equations in matrix form as follows:

\[ \begin{bmatrix} 3 & -2 & 1 \\ \lambda & -14 & 15 \\ 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Set Up the Determinant
For the system to have a non-trivial solution, the determinant of the coefficient matrix must be equal to zero. Thus, we need to calculate the determinant:

\[ \text{Det} = \begin{vmatrix} 3 & -2 & 1 \\ \lambda & -14 & 15 \\ 1 & 2 & 3 \end{vmatrix} \]

### Step 3: Calculate the Determinant
Using the determinant formula for a 3x3 matrix, we compute:

\[ \text{Det} = 3 \begin{vmatrix} -14 & 15 \\ 2 & 3 \end{vmatrix} - (-2) \begin{vmatrix} \lambda & 15 \\ 1 & 3 \end{vmatrix} + 1 \begin{vmatrix} \lambda & -14 \\ 1 & 2 \end{vmatrix} \]

Calculating these 2x2 determinants:

1. \( \begin{vmatrix} -14 & 15 \\ 2 & 3 \end{vmatrix} = (-14)(3) - (15)(2) = -42 - 30 = -72 \)

2. \( \begin{vmatrix} \lambda & 15 \\ 1 & 3 \end{vmatrix} = (\lambda)(3) - (15)(1) = 3\lambda - 15 \)

3. \( \begin{vmatrix} \lambda & -14 \\ 1 & 2 \end{vmatrix} = (\lambda)(2) - (-14)(1) = 2\lambda + 14 \)

Substituting these back into the determinant:

\[ \text{Det} = 3(-72) + 2(3\lambda - 15) + (2\lambda + 14) \]

### Step 4: Simplify the Determinant
Now simplify:

\[ \text{Det} = -216 + 6\lambda - 30 + 2\lambda + 14 \]
\[ = -216 - 30 + 14 + 8\lambda \]
\[ = -232 + 8\lambda \]

### Step 5: Set the Determinant to Zero
For a non-trivial solution, we set the determinant equal to zero:

\[ -232 + 8\lambda = 0 \]

### Step 6: Solve for \( \lambda \)
Solving for \( \lambda \):

\[ 8\lambda = 232 \]
\[ \lambda = \frac{232}{8} = 29 \]

### Step 7: Calculate \( \lambda^2 \)
Now, we find \( \lambda^2 \):

\[ \lambda^2 = 29^2 = 841 \]

### Final Answer
Thus, the value of \( \lambda^2 \) is \( \boxed{841} \).