If the following three linear equations have a non-trivial solution , then
x + 4ay + az = 0
x + 3by + bz = 0
x + 2cy + cz = 0

Options

A. a , b, c are in A.P.
B. a, b, c are in G.P.
C. a, b, c are in H.P.
D. a + b + c = 0

Answer

To determine if the given linear equations have a non-trivial solution, we need to analyze the system of equations:

1. \( x + 4ay + az = 0 \)
2. \( x + 3by + bz = 0 \)
3. \( x + 2cy + cz = 0 \)

### Step 1: Write the system in matrix form
We can express the system of equations in matrix form \( A \mathbf{x} = 0 \), where \( A \) is the coefficient matrix and \( \mathbf{x} \) is the vector of variables.

The coefficient matrix \( A \) is:
\[ A = \begin{bmatrix} 1 & 4a & a \\ 1 & 3b & b \\ 1 & 2c & c \end{bmatrix} \]

### Step 2: Set up the determinant
For the system to have a non-trivial solution, the determinant of the coefficient matrix must be zero:
\[ \text{det}(A) = 0 \]

### Step 3: Calculate the determinant
We can calculate the determinant of matrix \( A \) using the formula for the determinant of a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(dg - eh) + c(dh - eg) \]
For our matrix, this becomes:
\[ \text{det}(A) = 1 \cdot (3b \cdot c - 2c \cdot b) - 4a \cdot (1 \cdot c - 1 \cdot b) + a \cdot (1 \cdot 2c - 1 \cdot 3b) \]

Calculating each term:
1. \( 3bc - 2bc = bc \)
2. \( 4a(c - b) \)
3. \( 2ac - 3ab \)

Putting it all together:
\[ \text{det}(A) = 1 \cdot bc - 4a(c - b) + a(2c - 3b) \]
This simplifies to:
\[ bc - 4ac + 4ab + 2ac - 3ab = 0 \]
Combining like terms:
\[ bc - 2ac + ab = 0 \]

### Step 4: Factor the determinant equation
We can rearrange the equation:
\[ bc = 2ac - ab \]
Dividing through by \( abc \) (assuming \( a, b, c \neq 0 \)):
\[ \frac{bc}{abc} = \frac{2ac}{abc} - \frac{ab}{abc} \]
This simplifies to:
\[ \frac{1}{a} - \frac{2}{b} + \frac{1}{c} = 0 \]

### Step 5: Rearranging the equation
Rearranging gives us:
\[ \frac{1}{a} - \frac{1}{b} = \frac{1}{b} - \frac{1}{c} \]
This implies that \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in arithmetic progression.

### Step 6: Conclusion
Since \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in arithmetic progression, it follows that \( a, b, c \) are in harmonic progression.

Thus, the final conclusion is:
**The values \( a, b, c \) are in harmonic progression.**