Determinants Question 5
Question 5 - 2024 (29 Jan Shift 2)
Let for any three distinct consecutive terms $\mathrm{a}, \mathrm{b}, \mathrm{c}$ 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 $(\mathrm{PQ})^{2}$ is equal to
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Answer (113)
Solution
$\because a, b, c$ and in A.P
$\Rightarrow 2 b=a+c \Rightarrow a-2 b+c=0$
$\therefore a x+b y+c$ passes through fixed point $(1,-2)$
$\therefore P=(1,-2)$
For infinite solution,
$\mathrm{D}=\mathrm{D} 1=\mathrm{D} 2=\mathrm{D} 3=0$
$D:\left|\begin{array}{lll}1 & 1 & 1 \ 2 & 5 & \alpha \ 1 & 2 & 3\end{array}\right|=0$
$\Rightarrow \alpha=8$
$D_{1}:\left|\begin{array}{lll}6 & 1 & 1 \ \beta & 5 & \alpha \ 4 & 2 & 3\end{array}\right|=0 \Rightarrow \beta=6$
$\therefore Q=(8,6)$
$\therefore P Q^{2}=113$
