a,b,c are the sides of a triangle ABC. If the lines `ax+by+c=0,bx+cy+a=0` and `cx+ay+b=0` be concurrent then `Delta ABC` is

Options

A. right angled
B. isosceles
C. equilateral
D. none

Answer

To solve the problem, we need to determine the condition under which the lines given by the equations \( ax + by + c = 0 \), \( bx + cy + a = 0 \), and \( cx + ay + b = 0 \) are concurrent.

### Step-by-step Solution:

1. **Understanding Concurrency**:
For three lines to be concurrent, the determinant formed by their coefficients must be equal to zero. The lines can be represented in the form \( A_1x + B_1y + C_1 = 0 \), \( A_2x + B_2y + C_2 = 0 \), and \( A_3x + B_3y + C_3 = 0 \).

2. **Setting Up the Determinant**:
The determinant for the lines is given by:
\[ \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0 \]

3. **Calculating the Determinant**:
We can expand this determinant using 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 the 2x2 determinants:
- \( \begin{vmatrix} c & a \\ a & b \end{vmatrix} = cb - a^2 \)
- \( \begin{vmatrix} b & a \\ c & b \end{vmatrix} = bb - ac = b^2 - ac \)
- \( \begin{vmatrix} b & c \\ c & a \end{vmatrix} = ba - c^2 \)

Substituting back, we get:
\[ a(cb - a^2) - b(b^2 - ac) + c(ba - c^2) = 0 \]

4. **Expanding the Expression**:
Expanding this gives:
\[ acb - a^3 - b^3 + abc + abc - c^3 = 0 \]
Simplifying, we have:
\[ 3abc - (a^3 + b^3 + c^3) = 0 \]

5. **Using the Identity**:
We can use the identity \( a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc) \). Setting this equal to zero gives us two cases:
- \( a + b + c = 0 \) (not possible for triangle sides)
- \( a^2 + b^2 + c^2 - ab - ac - bc = 0 \)

6. **Conclusion**:
The second condition simplifies to:
\[ (a - b)^2 + (b - c)^2 + (c - a)^2 = 0 \]
This implies that \( a = b = c \). Thus, triangle ABC is an equilateral triangle.

### Final Answer:
The triangle ABC is an **equilateral triangle**.