Let for any three distinct consecutive terms a , b , c of an A.P, the lines a x + b y + c = 0 be concurrent…

Given,

$a,b,c$ are in $\mathrm{A}.\mathrm{P}.$

$\Rightarrow 2b=a+c\Rightarrow a-2b+c=0$

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

 The line $ax+by+c=0$ passes through fixed point $\left(1,-2\right)$

$\therefore P=(1,-2)$

Now, we know that for infinite solution,

$D=D_1=D_2=D_3=0$

$D=\left|\begin{array}{lll}1& 1& 1\\ 2& 5& \alpha \\ 1& 2& 3\end{array}\right|=0$

$\Rightarrow \alpha =8$

And $D_1=\left|\begin{array}{lll}6& 1& 1\\ \beta & 5& \alpha \\ 4& 2& 3\end{array}\right|=0\Rightarrow \beta =6$

So, the point $Q=\left(\alpha ,\beta \right)=(8,6)$

Hence, $P{Q}^2=7^2+8^2=49+64=113$