Question

Let for any three distinct consecutive terms \( a, b, c \) of an A.P., the lines \( ax + by + c = 0 \) be concurrent at the point \( P \) and \( Q (\alpha, \beta) \) be a point such that the system of equations \[x + y + z = 6,\]\[2x + 5y + \alpha z = \beta,\]\[x + 2y + 3z = 4,\]has infinitely many solutions. Then \( (PQ)^2 \) is equal to ______.

Answer 1

Correct Answer: 113

Since \( a, b, c \) are in A.P., we have:  \(2b = a + c \implies a - 2b + c = 0\)

This implies that the line \( ax + by + c = 0 \) passes through the fixed point \( (1, -2) \). Therefore, \( P = (1, -2) \).  

For the system of equations to have infinitely many solutions, the determinants \( D = D_1 = D_2 = D_3 = 0 \) must hold.  

Step 1. Calculate \( a \) using \( D = 0 \): 
 
  \(D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & a \\ 1 & 2 & 3 \end{vmatrix} = 0\)
 
  Expanding this determinant, we get:  
  \(a = 8\)
Step 2. Calculate \( b \) using \( D_1 = 0 \):
 
  \(D_1 = \begin{vmatrix} 6 & 1 & 1 \\ 4 & 2 & 3 \\ \beta & 5 & a \end{vmatrix} = 0\)
 
  Substituting \( a = 8 \):  
   \(\beta = 6\)
Thus, the point \( Q = (8, 6) \).  

Step 3. Calculate \( (PQ)^2 \):
  \((PQ)^2 = (8 - 1)^2 + (6 - (-2))^2\)
  
  \(= 7^2 + 8^2 = 49 + 64 = 113\)
The Correct Answer is: \( PQ^2 = 113 \)

Answer 2

Given that \( a \), \( b \), and \( c \) are in Arithmetic Progression (A.P.), we have the relation: \[ 2b = a + c \quad \text{or} \quad a - 2b + c = 0 \] From this, the line \( ax + by + c = 0 \) passes through the fixed point \( P(1, -2) \), meaning the point \( P \) lies on this line. For the system of equations to have infinite solutions, we set up the system as: \[ \text{D} = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{vmatrix} = 0 \] Expanding the determinant: \[ \text{D} = 1 \times \begin{vmatrix} 5 & \alpha \\ 2 & 3 \end{vmatrix} - 1 \times \begin{vmatrix} 2 & \alpha \\ 1 & 3 \end{vmatrix} + 1 \times \begin{vmatrix} 2 & 5 \\ 1 & 2 \end{vmatrix} \] After simplifying, we find that: \[ \alpha = 8 \] Next, for the second system to have infinite solutions, we solve the following determinant: \[ \text{D1} = \begin{vmatrix} 6 & 1 & 1 \\ \beta & 5 & \alpha \\ 4 & 2 & 3 \end{vmatrix} = 0 \] Substituting \( \alpha = 8 \), we solve for \( \beta \) and find: \[ \beta = 6 \] Thus, the point \( Q \) has coordinates \( Q(8, 6) \). Finally, we calculate the square of the distance \( (PQ)^2 \) between the points \( P(1, -2) \) and \( Q(8, 6) \): \[ (PQ)^2 = (8 - 1)^2 + (6 - (-2))^2 = 7^2 + 8^2 = 49 + 64 = 113 \]

Final Answer:

The value of \( (PQ)^2 \) is: \[ \boxed{113} \]