Let the system of equations :
$\begin{aligned}
& 2 x+3 y+5 z=9 \\ & 7 x+3 y-2 z=8 \\ & 12 x+3 y-(4+\lambda) z=16-\mu
\end{aligned}$
have infinitely many solutions. Then the radius of the circle centred at $(\lambda, \mu)$ and touching the line $4 x=3 y$ is

Question

Aicharya · Accepted Answer

$\begin{aligned} & \left|\begin{array}{ccc}2 & 3 & 5 \ 7 & 3 & -2 \ 12 & 3 & -(\lambda+4)\end{array}ight|=0 \ & \Rightarrow 12(-21)-3(-39)-(\lambda+4)(-15)=0 \ & \Rightarrow-252+117+15(1+4)=0 \ & \Rightarrow 15 \lambda+177-252=0 \ & \Rightarrow 15 \lambda-75=0 \Rightarrow \lambda=5 \ & \left|\begin{array}{ccc}9 & 3 & 5 \ 8 & 3 & -2 \ 16-\mu & 3 & -9\end{array}ight|=0 \Rightarrow\left|\begin{array}{ccc}1 & 0 & 7 \ \mu-8 & 0 & 7 \ 16-\mu & 3 & -9\end{array}ight|=0 \ & \Rightarrow 7-7(\mu-8)=0 \Rightarrow 1-(\mu-8)=0 \Rightarrow \mu=9 \ & \Rightarrow 	ext { centre of circle }(5,9) \ & 	ext { radius }=	ext { length of } \perp 	ext { from centre }(5, \quad 9)= \ & \left|\frac{20-27}{5}ight|=\frac{7}{5}\end{aligned}$

Let the system of equations : $\begin{aligned} & 2 x+3 y+5 z=9 \\ & 7 x+3 y-2 z=8 \\ & 12 x+3 y-(4+\lambda)…

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