If the equations
`(b + c) x+(c+a) y +(a+b)z =0`,
`cx+ay+bz=0`,
`ax+by+cz=0 `
are consistent with more than one solution, then

Options

A. `a+b+c=0`
B. ` a=b=c`
C. `a+b+c=2`
D. `2a=3b=4c`

Answer

To determine the conditions under which the given system of equations is consistent with more than one solution, we start by analyzing the equations:

1. \((b + c)x + (c + a)y + (a + b)z = 0\)
2. \(cx + ay + bz = 0\)
3. \(ax + by + cz = 0\)

### Step 1: Identify the Coefficient Matrix
We can express the system of equations in matrix form as follows:

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

Let \(A\) be the coefficient matrix:

\[ A = \begin{bmatrix} b + c & c + a & a + b \\ c & a & b \\ a & b & c \end{bmatrix} \]

### Step 2: Condition for More than One Solution
For the system to have more than one solution, the determinant of the coefficient matrix \(A\) must be zero:

\[ \text{det}(A) = 0 \]

### Step 3: Calculate the Determinant
We will calculate the determinant of matrix \(A\):

\[ \text{det}(A) = \begin{vmatrix} b + c & c + a & a + b \\ c & a & b \\ a & b & c \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix, we expand it:

\[ \text{det}(A) = (b+c) \begin{vmatrix} a & b \\ b & c \end{vmatrix} - (c+a) \begin{vmatrix} c & b \\ a & c \end{vmatrix} + (a+b) \begin{vmatrix} c & a \\ a & b \end{vmatrix} \]

Calculating each of these 2x2 determinants:

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

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

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

Substituting these back into the determinant expression:

\[ \text{det}(A) = (b+c)(ac - b^2) - (c+a)(c^2 - ab) + (a+b)(cb - a^2) \]

### Step 4: Simplify the Determinant
Now we simplify the expression obtained. After performing the necessary algebraic manipulations, we will arrive at a condition involving \(a\), \(b\), and \(c\).

The determinant simplifies to:

\[ \text{det}(A) = a^3 + b^3 + c^3 - 3abc \]

### Step 5: Set the Determinant to Zero
For the system to have more than one solution, we set the determinant to zero:

\[ a^3 + b^3 + c^3 - 3abc = 0 \]

This can be factored using the identity:

\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

### Step 6: Conclusion
Thus, the conditions for the system of equations to be consistent with more than one solution are:

1. \(a + b + c = 0\) or
2. \(a^2 + b^2 + c^2 - ab - ac - bc = 0\) (which implies \(a = b = c\))