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Let for any three distinct consecutive terms $a, b, c$ of an $A.P,$ the lines $a x+b y+c=0$ be concurrent at the point $\mathrm{P}$ and $\mathrm{Q}(\alpha, \beta)$ be a point such that the system of equations $ x+y+z=6, $ $ 2 x+5 y+\alpha z=\beta$ and $x+2 y+3 z=4$, has infinitely many solutions. Then $(P Q)^2$ is equal to________.
$123$
$113$
$421$
$131$
Solution
$\because \mathrm{a}, \mathrm{b}, \mathrm{c}$ and in $A.P$
$\Rightarrow 2 \mathrm{~b}=\mathrm{a}+\mathrm{c} \Rightarrow \mathrm{a}-2 \mathrm{~b}+\mathrm{c}=0$
$\therefore \mathrm{ax}+\mathrm{by}+\mathrm{c}$ passes through fixed point $(1,-2)$
$\therefore \mathrm{P}=(1,-2)$
For infinite solution,
$\mathrm{D}=\mathrm{D}_1=\mathrm{D}_2=\mathrm{D}_3=0$
$\mathrm{D}:\left|\begin{array}{lll}1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3\end{array}\right|=0$
$\Rightarrow \alpha=8$
$D_1:\left|\begin{array}{lll}6 & 1 & 1 \\ \beta & 5 & \alpha \\ 4 & 2 & 3\end{array}\right|=0 \Rightarrow \beta=6$
$ \therefore Q=(8,6) $
$ \therefore P Q^2=113$
