JEE MAIN - Mathematics (2024 - 29th January Evening Shift - No. 26)
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 $$P$$ and $$Q(\alpha, \beta)$$ be a point such that the system of equations
$$\begin{aligned} & x+y+z=6, \\ & 2 x+5 y+\alpha z=\beta \text { and } \end{aligned}$$
$$x+2 y+3 z=4$$, has infinitely many solutions. Then $$(P Q)^2$$ is equal to _________.
Explanation
$$\because \mathrm{a}, \mathrm{b}, \mathrm{c}$$ and in A.P
$$\Rightarrow 2 b=a+c \Rightarrow a-2 b+c=0$$
$$\therefore \mathrm{ax}+\mathrm{by}+\mathrm{c}$$ passes through fixed point $$(1,-2)$$
$$\therefore \mathrm{P}=(1,-2)$$
For infinite solution,
$$\begin{aligned} & D=D_1=D_2=D_3=0 \\ & D:\left|\begin{array}{lll} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{array}\right|=0 \\ & \Rightarrow \alpha=8 \\ & D_1:\left|\begin{array}{lll} 6 & 1 & 1 \\ \beta & 5 & \alpha \\ 4 & 2 & 3 \end{array}\right|=0 \Rightarrow \beta=6 \\ & \therefore Q=(8,6) \\ & \therefore Q^2=113 \end{aligned}$$
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