Let for any three distinct consecutive terms a, b, c of an A.P, the lines ax + by + c =0 be concurrent at the point P and Q (α, β) be a point such that the system of equations x+y+z=6 , 2 x+5 y+α z=β and x +2 y +3 z =4 , has infinitely many solutions. Then ( PQ ) 2 is equal to \ \ \ \ \ \ \ .

Mathematics · Determinants

JEE Main 2024 — 29 January, Shift 2 — Question 22

Let for any three distinct consecutive terms $a, b, c$ of an A.P, the lines $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ be concurrent at the point 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 $\mathrm{x}+2 \mathrm{y}+3 \mathrm{z}=4$ , has infinitely many solutions. Then $(\mathrm{PQ})^{2}$ is equal to $\_\_\_\_\_\_\_$ .

Answer: 113

Numerical answer — enter this value.

Step-by-step 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

D:\left| \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{array} \right|=0 \Rightarrow \alpha=8

D_{1}:\left| \begin{array}{ccc} 6 & 1 & 1 \\ \beta & 5 & \alpha \\ 4 & 2 & 3 \end{array} \right|=0 \Rightarrow \beta=6

\therefore Q=(8,6)

\therefore PQ^{2}=113

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Mathematics
Chapter: Determinants
Topic: Consistency of Non-homogeneous system