If the system of linear equations
`2x+2ay+az=0`
`2x+3by+bz=0`
`2x+4cy+cz=0`
where a,b,c `in` R are non - zero and distinct , has a non-zero solution, then :

Options

A. `a+b+c=0`
B. a,b,c are in A.P.
C. `1/a,1/b,1/c` are in A.P.
D. a,b,c are in G.P.

Answer

To solve the given system of linear equations, we need to find the condition under which the system has a non-zero solution. The equations are:

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

We can express this system in matrix form as follows:

\[ \begin{bmatrix} 2 & 2a & a \\ 2 & 3b & b \\ 2 & 4c & c \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

For the system to have a non-zero solution, the determinant of the coefficient matrix must be zero. Let's denote the determinant as \( D \):

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

### Step 1: Calculate the determinant

We can use the property of determinants that allows us to simplify rows. Subtract the first row from the second and third rows:

\[ D = \begin{vmatrix} 2 & 2a & a \\ 0 & 3b - 2a & b - a \\ 0 & 4c - 2a & c - a \end{vmatrix} \]

### Step 2: Expand the determinant

Now, we can expand the determinant along the first column:

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

### Step 3: Calculate the 2x2 determinant

Now we calculate the 2x2 determinant:

\[ D = 2 \left( (3b - 2a)(c - a) - (b - a)(4c - 2a) \right) \]

### Step 4: Simplify the expression

Expanding this gives:

\[ D = 2 \left( 3bc - 3ba - 2ac + 2a^2 - (4bc - 2ab - 4ac + 2a^2) \right) \]

Combining like terms:

\[ D = 2 \left( 3bc - 4bc + 2ab - 2ac + 2a^2 - 2a^2 \right) \]

This simplifies to:

\[ D = 2 \left( -bc + 2ab - 2ac \right) \]

### Step 5: Set the determinant to zero

For the system to have a non-zero solution, we set \( D = 0 \):

\[ -bc + 2ab - 2ac = 0 \]

### Step 6: Rearranging the equation

Rearranging gives us:

\[ 2ab - 2ac = bc \]

Dividing through by \( b \) (since \( b \neq 0 \)):

\[ 2a - 2c = \frac{bc}{b} \]

This leads us to:

\[ 2a = 2c + c \]

### Step 7: Conclusion

From the above, we can conclude that:

\[ \frac{1}{a} + \frac{1}{c} = \frac{2}{b} \]

This indicates that \( a, b, c \) are in Harmonic Progression.