The system of linear equations
x + y + z = 0
`(2x)/(a) + (3y)/(b) + (4z)/(c ) = 0`
`(x)/(a) + (y)/(b) + (z)/(c ) = 0`
has non trivia solution then

Options

A. a + b + c = 0
B. a , b , c are in GP
C. `1/(a),1/(b),1/(c)` are in AP
D. `a,b,c ` are in AP

Explanation

To determine the conditions under which the system of linear equations has a non-trivial solution, we need to analyze the given equations:

1. \( x + y + z = 0 \)
2. \( \frac{2x}{a} + \frac{3y}{b} + \frac{4z}{c} = 0 \)
3. \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 0 \)

### Step 1: Form the Coefficient Matrix
To find the conditions for a non-trivial solution, we need to set up the coefficient matrix from the equations. The coefficients of \(x\), \(y\), and \(z\) from the three equations can be represented as follows:

\[ \begin{bmatrix} 1 & 1 & 1 \\ \frac{2}{a} & \frac{3}{b} & \frac{4}{c} \\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c} \end{bmatrix} \]

### Step 2: Calculate the Determinant
For the system to have a non-trivial solution, the determinant of this matrix must be zero. We will denote the determinant as \(D\):

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ \frac{2}{a} & \frac{3}{b} & \frac{4}{c} \\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c} \end{vmatrix} \]

### Step 3: Expand the Determinant
Using the determinant formula, we can expand \(D\):

\[ D = 1 \cdot \left( \frac{3}{b} \cdot \frac{1}{c} - \frac{4}{c} \cdot \frac{1}{b} \right) - 1 \cdot \left( \frac{2}{a} \cdot \frac{1}{c} - \frac{4}{c} \cdot \frac{1}{a} \right) + 1 \cdot \left( \frac{2}{a} \cdot \frac{1}{b} - \frac{3}{b} \cdot \frac{1}{a} \right) \]

This simplifies to:

\[ D = \frac{3}{bc} - \frac{4}{bc} - \left( \frac{2}{ac} - \frac{4}{ac} \right) + \left( \frac{2}{ab} - \frac{3}{ab} \right) \]

### Step 4: Simplify the Expression
Combining like terms gives:

\[ D = \frac{-1}{bc} + \frac{2}{ac} - \frac{1}{ab} \]

### Step 5: Set the Determinant to Zero
For a non-trivial solution, we set \(D = 0\):

\[ \frac{-1}{bc} + \frac{2}{ac} - \frac{1}{ab} = 0 \]

### Step 6: Multiply by \(abc\) to Eliminate Denominators
Multiplying through by \(abc\) to eliminate the denominators:

\[ -a + 2b - c = 0 \]

### Step 7: Rearranging the Equation
Rearranging gives us the relationship:

\[ c + a = 2b \]

### Step 8: Conclusion
This implies that \(a\), \(b\), and \(c\) are in arithmetic progression (AP).

### Final Answer
Thus, the condition for the system of equations to have a non-trivial solution is that \(a\), \(b\), and \(c\) are in AP.