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Q. Let $p, q, r$ be nonzero real numbers that are, respectively, the $10^{\text {th }}, 100^{\text {th }}$ and $1000^{\text {th }}$ terms of a harmonic progression. Consider the system of linear equations
$x+y+z=1 $
$10 x+100 y+1000 z=0$
$q r x+p r y+p q z=0 .$
List I List II
A If $\frac{q}{r}=10$, then the system of linear equations has P $x =0, y =\frac{10}{9}, z =-\frac{1}{9}$ as a solution
B If $\frac{p}{r} \neq 100$, then the system of linear equations has Q $x =\frac{10}{9}, y =-\frac{1}{9}, z =0$ as a solution
C If $\frac{p}{q} \neq 10$, then the system of linear equations has R infinitely many solutions
D If $\frac{ p }{ q }=10$, then the system of linear S no solution
T at least one solution
The correct option is:

JEE AdvancedJEE Advanced 2022

Solution:

If $\frac{ q }{ r }=10 \Rightarrow A = D \Rightarrow D _{ x }= D _{ y }= D _{ z }=0$
So, there are infinitely many solutions Look of infinitely many solutions can be given as
$x+y+z=1$
$\& 10 x +100 y +1000 z =0 \Rightarrow x +10 y +100 z =0$
Let $z =\lambda$
then $x+y=1-\lambda$
and $x+10 y=-100 \lambda$
$\Rightarrow x =\frac{10}{9}+10 \lambda ; y =\frac{-1}{9}-11 \lambda$
i.e., $(x, y, z) \equiv\left(\frac{10}{9}+10 \lambda, \frac{-1}{9}-11 \lambda, \lambda\right)$
$Q \left(\frac{10}{9}, \frac{-1}{9}, 0\right)$ valid for $\lambda=0$
$P \left(0, \frac{10}{9}, \frac{-1}{9}\right)$ not valid for any $\lambda$.
(I) $\rightarrow$ Q,R,T
(II) If $\frac{p}{r} \neq 100$, then $D_y \neq 0$
So no solution
(II) $\rightarrow$ (S)
(III) If $\frac{ p }{ q } \neq 10$, then $D _2 \neq 0$ so, no solution
(III) $\rightarrow$ (S)
(IV) If $\frac{p}{q}=10 \Rightarrow D_z=0 \Rightarrow D_x=D_y=0$
so infinitely many solution
(IV) $\rightarrow$ Q, R, T