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 :

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Answer: A. \frac{7}{5} \begin{aligned} & \Delta=\left|\begin{array}{ccc} 2 & 3 & 5 \ 7 & 3 & -2 \ 12 & 3 & -(4+\lambda) \end{array}ight| \ & =2(-12-3 \lambda+6)-3(-28-7 \lambda+24)+5(21-36) \ & =-12-6 \lambda+12+21 \lambda-75 \ & =15 \lambda-75 \ & \Rightarrow 15 \lambda-75=0 \ & \Rightarrow \lambda=5 \ & \Delta_1=\left|\begin{array}{ccc} 9 & 3 & 5 \ 8 & 3 & -2 \ 16-\mu & 3 & -9 \end{array}ight| \ & =9(-27+6)-3(-72+32-2 \mu)+5(24-48+3 \mu) \ & =-189+120+6 \mu-120+15 \mu \ & =21 \mu-189=0 \ & \Rightarrow \mu=9 \end{aligned} \begin{aligned} & 	herefore r=\left|\frac{4(5)-3(9)}{\sqrt{(4)^2+(3)^2}}ight| \ & r=\frac{7}{5} \end{aligned}

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

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