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

Given that \( a, b, c \) are in A.P., the condition \(2b = a + c\) simplifies to \( a - 2b + c = 0 \). This indicates that the line \( ax + by + c = 0 \) passes through the fixed point \( (1, -2) \), thus \( P = (1, -2) \).

For the system of equations to possess infinitely many solutions, all determinants \( D, D_1, D_2, D_3 \) must be equal to zero.

Step 1. Determine \( a \) by setting \( D = 0 \):
\(D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & a \\ 1 & 2 & 3 \end{vmatrix} = 0\)
Expanding the determinant yields \(a = 8\).

Step 2. Determine \( b \) (represented by \(\beta\)) by setting \( D_1 = 0 \):
\(D_1 = \begin{vmatrix} 6 & 1 & 1 \\ 4 & 2 & 3 \\ \beta & 5 & a \end{vmatrix} = 0\)
Substituting \( a = 8 \) into the equation results in \(\beta = 6\). Therefore, 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 final result is \( PQ^2 = 113 \).