if the system of equations
`(a-t)x+by +cz=0`
`bx+(c-t) y+az=0`
`cx+ay+(b-t)z=0`
has non-trivial solutions then product of all possible values of t is

Options

A. `|{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}|`
B. `a+b+c`
C. `a^(2)+b^(2)+c^(2)`
D. `1`

Explanation

To solve the given problem, we need to find the product of all possible values of \( t \) for which the system of equations has non-trivial solutions. The system of equations is:

1. \( (a-t)x + by + cz = 0 \)
2. \( bx + (c-t)y + az = 0 \)
3. \( cx + ay + (b-t)z = 0 \)

### Step 1: Write the system in matrix form
The system can be represented in matrix form as follows:

\[ \begin{bmatrix} a-t & b & c \\ b & c-t & a \\ c & a & b-t \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Set the determinant to zero
For the system to have non-trivial solutions, the determinant of the coefficient matrix must be zero:

\[ \Delta = \begin{vmatrix} a-t & b & c \\ b & c-t & a \\ c & a & b-t \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
Using the determinant formula for a 3x3 matrix, we can expand it:

\[ \Delta = (a-t) \begin{vmatrix} c-t & a \\ a & b-t \end{vmatrix} - b \begin{vmatrix} b & a \\ c & b-t \end{vmatrix} + c \begin{vmatrix} b & c-t \\ c & a \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} c-t & a \\ a & b-t \end{vmatrix} = (c-t)(b-t) - a^2 \)

2. \( \begin{vmatrix} b & a \\ c & b-t \end{vmatrix} = b(b-t) - ac \)

3. \( \begin{vmatrix} b & c-t \\ c & a \end{vmatrix} = b \cdot a - c(c-t) \)

Substituting these back into the determinant expression gives:

\[ \Delta = (a-t)((c-t)(b-t) - a^2) - b(b(b-t) - ac) + c(ba - c(c-t)) \]

### Step 4: Simplify the determinant
After expanding and simplifying the expression, we will obtain a polynomial in \( t \):

\[ \Delta = t^3 + \alpha t^2 + \beta t + \gamma = 0 \]

### Step 5: Identify the product of the roots
The product of the roots of the polynomial \( t^3 + \alpha t^2 + \beta t + \gamma = 0 \) can be found using Vieta's formulas. The product of the roots \( t_1, t_2, t_3 \) is given by:

\[ t_1 t_2 t_3 = -\frac{\gamma}{1} \]

### Step 6: Calculate \( \gamma \)
From the determinant calculation, we find that:

\[ \gamma = abc \]

Thus, the product of all possible values of \( t \) is:

\[ t_1 t_2 t_3 = abc \]

### Final Answer
The product of all possible values of \( t \) is \( abc \).