Verified Solution • 2025–26 Session moderate

QJEEM24Maebf1: JEE Main 2024 Mathematics - Class 12 JEEM24Maebf1 Solution: Let for any three distinct consecutive terms of an A.P, the…

Let for any three distinct consecutive terms a,b,ca, b, c of an A.P, the lines ax+by+c=0a x+b y+c=0 be concurrent at the point PP and Q(α,β)Q(\alpha, \beta) be a point such that the system of equations

amp;x+y+z=6,amp;2x+5y+αz=β and \begin{aligned}& x+y+z=6, \\& 2 x+5 y+\alpha z=\beta \text { and }\end{aligned}

x+2y+3z=4x+2 y+3 z=4 , has infinitely many solutions. Then (PQ)2(P Q)^2 is equal to _________.

Step-by-Step Verified Solution

∵a,b,c\because \mathrm{a}, \mathrm{b}, \mathrm{c} and in A.P

⇒2b=a+c⇒a−2b+c=0\Rightarrow 2 b=a+c \Rightarrow a-2 b+c=0

∴ax+by+c\therefore \mathrm{ax}+\mathrm{by}+\mathrm{c} passes through fixed point (1,−2)(1,-2)

∴P=(1,−2)\therefore \mathrm{P}=(1,-2)

For infinite solution,

amp;D=D1=D2=D3=0amp;D:∣1amp;1amp;12amp;5amp;α1amp;2amp;3∣=0amp;⇒α=8amp;D1:∣6amp;1amp;1βamp;5amp;α4amp;2amp;3∣=0⇒β=6amp;∴Q=(8,6)amp;∴Q2=113\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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