Let for any three distinct consecutive terms a, b, c of an A.P, the...

Given,

$a, b, c$ are in A . P .

$⇒ 2b = a + c ⇒ a - 2b + c = 0$

Now, on comparing above equation with $ax + by + c = 0$ we get,

 The line $ax + by + c = 0$ passes through fixed point \left(1, - 2\right)$</p><p>$∴ P = ( 1, - 2 )$</p><p>Now, we know that for infinite solution,</p><p>$D = {D}_{1} = {D}_{2} = {D}_{3} = 0$</p><p>$D = \left|$\begin{cases}1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3\end{cases}$\right| = 0$</p><p>$⇒ \alpha = 8$</p><p>And${D}_{1} = \left|$\begin{cases}6 & 1 & 1 \\ \beta & 5 & \alpha \\ 4 & 2 & 3\end{cases}$\right| = 0 ⇒ \beta = 6$</p><p>So, the point$Q = \left(\alpha, \beta\right) = ( 8, 6 )$</p><p>Hence,$P{Q}^{2} = {7}^{2} + {8}^{2} = 49 + 64 = 113$