The three lines `ax+by+c=0`,
`bx+cy+a=0,cx+ay+b=0`
are concurrent only when

Options

A. `a+b+c=0`
B. `a^(2)+b^(2)+c^(2)=ab+bc+ca`
C. `a^(3)+b^(3)+c^(3)=3ab`
D. None of these

Answer

To determine the condition under which the three lines \( ax + by + c = 0 \), \( bx + cy + a = 0 \), and \( cx + ay + b = 0 \) are concurrent, we will use the concept of determinants.

### Step-by-Step Solution:

1. **Write the equations of the lines in standard form**:
We have the three lines:
\[ L_1: ax + by + c = 0 \]
\[ L_2: bx + cy + a = 0 \]
\[ L_3: cx + ay + b = 0 \]

2. **Set up the determinant for concurrency**:
The lines are concurrent if the determinant of the coefficients of \( x \), \( y \), and the constant terms is zero. The determinant can be set up as follows:
\[ \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0 \]

3. **Calculate the determinant**:
Expanding the determinant, we can use the first row:
\[ = a \begin{vmatrix} c & a \\ a & b \end{vmatrix} - b \begin{vmatrix} b & a \\ c & b \end{vmatrix} + c \begin{vmatrix} b & c \\ c & a \end{vmatrix} \]

Calculating each of these 2x2 determinants:
- For the first determinant:
\[ = a (cb - a^2) \]
- For the second determinant:
\[ = -b (b^2 - ac) \]
- For the third determinant:
\[ = c (ba - c^2) \]

Putting it all together:
\[ a(cb - a^2) - b(b^2 - ac) + c(ba - c^2) = 0 \]

4. **Simplify the expression**:
Expanding this gives:
\[ acb - a^3 - b^3 + abc + abc - c^3 = 0 \]
Combining like terms:
\[ 3abc - (a^3 + b^3 + c^3) = 0 \]

5. **Rearranging the equation**:
Thus, we have:
\[ a^3 + b^3 + c^3 = 3abc \]

6. **Conclusion**:
The condition for the three lines to be concurrent is:
\[ a^3 + b^3 + c^3 - 3abc = 0 \]