The lines `px +qy+r=0, qx + ry + p =0,rx + py+q=0,` are concurrant then

Options

A. p+q+r=0
B. `p^(2) + q^(2) + r^(2) = pr+rp+pq`
C. `p^(3) + q^(3) + r^(3) = 3pqr`
D. none of these

Explanation

To determine the condition for the lines \( px + qy + r = 0 \), \( qx + ry + p = 0 \), and \( rx + py + q = 0 \) to be concurrent, we can use the concept of determinants. The lines are concurrent if the determinant of the coefficients of \( x \), \( y \), and the constant terms is equal to zero.

### Step-by-Step Solution:

1. **Write the equations in standard form**:
The three equations given are:
\[ \text{(1)} \quad px + qy + r = 0 \]
\[ \text{(2)} \quad qx + ry + p = 0 \]
\[ \text{(3)} \quad rx + py + q = 0 \]

2. **Set up the determinant**:
The determinant of the coefficients of \( x \), \( y \), and the constant terms is given by:
\[ D = \begin{vmatrix} p & q & r \\ q & r & p \\ r & p & q \end{vmatrix} \]

3. **Calculate the determinant**:
We can expand this determinant using the rule of Sarrus or cofactor expansion. The determinant can be calculated as follows:
\[ D = p \begin{vmatrix} r & p \\ p & q \end{vmatrix} - q \begin{vmatrix} q & p \\ r & q \end{vmatrix} + r \begin{vmatrix} q & r \\ r & p \end{vmatrix} \]

Calculating the 2x2 determinants:
\[ \begin{vmatrix} r & p \\ p & q \end{vmatrix} = rq - p^2 \]
\[ \begin{vmatrix} q & p \\ r & q \end{vmatrix} = q^2 - rp \]
\[ \begin{vmatrix} q & r \\ r & p \end{vmatrix} = qp - r^2 \]

Plugging these back into the determinant:
\[ D = p(rq - p^2) - q(q^2 - rp) + r(qp - r^2) \]

4. **Simplify the expression**:
Expanding this gives:
\[ D = prq - p^3 - q^3 + qrp + rqp - r^3 \]
Combining like terms:
\[ D = 3pqr - (p^3 + q^3 + r^3) \]

5. **Set the determinant to zero**:
For the lines to be concurrent, we need:
\[ 3pqr - (p^3 + q^3 + r^3) = 0 \]
Rearranging gives:
\[ p^3 + q^3 + r^3 = 3pqr \]

### Conclusion:
The condition for the lines to be concurrent is:
\[ p^3 + q^3 + r^3 = 3pqr \]