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 solve the system of equations for the value of \( \lambda^2 \) such that the system has a non-trivial solution, we will use the determinant of the coefficients matrix. The equations given are:

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

### Step 1: Write the system in matrix form
We can express 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 zero:

\[ \Delta = \begin{vmatrix} 3 & -2 & 1 \\ \lambda & -14 & 15 \\ 1 & 2 & 3 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
We will calculate the determinant using the rule of Sarrus or cofactor expansion. The determinant can be calculated as follows:

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

Calculating each of the 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:

\[ \Delta = 3(-72) + 2(3\lambda - 15) + (2\lambda + 14) \]

### Step 4: Simplify the determinant
Now, simplify the expression:

\[ \Delta = -216 + 6\lambda - 30 + 2\lambda + 14 \]
\[ \Delta = -216 - 30 + 14 + 6\lambda + 2\lambda \]
\[ \Delta = -232 + 8\lambda \]

### Step 5: Set the determinant to zero
Set the determinant equal to zero for a non-trivial solution:

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

### Step 6: Solve for \( \lambda \)
Rearranging gives:

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

### Step 7: Find \( \lambda^2 \)
Now, we need to find \( \lambda^2 \):

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

### Final Answer
Thus, the value of \( \lambda^2 \) is:

\[ \boxed{841} \]