If the lines `p_(r)x+q_(r)y+1=0,r=1,2,3` be concurrent then the points `(p_(r),q_(r)),r=1,2,3` are collinear.

Answer

To determine if the points \((p_r, q_r)\) for \(r = 1, 2, 3\) are collinear when the lines \(p_r x + q_r y + 1 = 0\) are concurrent, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Concurrent Lines**:
The lines \(p_1 x + q_1 y + 1 = 0\), \(p_2 x + q_2 y + 1 = 0\), and \(p_3 x + q_3 y + 1 = 0\) are said to be concurrent if they meet at a single point.

2. **Setting Up the Determinant**:
For the lines to be concurrent, the determinant formed by their coefficients must equal zero. The determinant can be set up as follows:
\[ \begin{vmatrix} p_1 & q_1 & -1 \\ p_2 & q_2 & -1 \\ p_3 & q_3 & -1 \end{vmatrix} = 0 \]

3. **Calculating the Determinant**:
The determinant can be expanded as:
\[ p_1 \begin{vmatrix} q_2 & -1 \\ q_3 & -1 \end{vmatrix} - q_1 \begin{vmatrix} p_2 & -1 \\ p_3 & -1 \end{vmatrix} + (-1) \begin{vmatrix} p_2 & q_2 \\ p_3 & q_3 \end{vmatrix} \]
This simplifies to:
\[ p_1(q_2(-1) - q_3(-1)) - q_1(p_2(-1) - p_3(-1)) - (p_2 q_3 - p_3 q_2) = 0 \]
Which can be rearranged to:
\[ p_1(q_3 - q_2) + q_1(p_3 - p_2) + (p_2 q_3 - p_3 q_2) = 0 \]

4. **Interpreting the Result**:
The above equation represents the condition for the points \((p_1, q_1)\), \((p_2, q_2)\), and \((p_3, q_3)\) to be collinear. When the determinant is zero, it indicates that the area of the triangle formed by these points is zero, which means that they lie on a straight line.

5. **Conclusion**:
Since the condition for the lines to be concurrent leads to a determinant that indicates collinearity of the points, we conclude that if the lines \(p_r x + q_r y + 1 = 0\) are concurrent, then the points \((p_r, q_r)\) for \(r = 1, 2, 3\) are indeed collinear.

### Final Answer:
Thus, the statement is **True**.