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

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: $PQ^2 = 113$