The equation of the line passing the point of intersection of the lines 3x+2y+4=0, 2x+5y-1=0 and whose distance from the point (2, -1) is 2 is

$\begin{array}{l}GivenEquation3x+2y+4=0--\left(1\right)\\ 2x+5y-1=0---\left(2\right)\\ Solving\left(1\right)\&\left(2\right)\\ x y 1\\ \begin{array}{cccc}2& 4& 3& 2\\ 5& -1& 2& 5\end{array}\\ \Rightarrow \frac{x}{-2-20}=\frac{y}{8+3}=\frac{1}{15-4}\\ \Rightarrow \frac{x}{-22}=\frac{y}{11}=\frac{1}{11}\\ \therefore (x,y)=(-2,1)\\ Equationoflinepas\sin gthrough(-2,1)is\\ y-1=m(x+2)--\left(3\right)\\ \Rightarrow mx-y+2m+1=0\end{array}$

$\begin{array}{l}Givendis\tan cefrom(2,-1)to\left(3\right)=2\\ \Rightarrow \frac{|2m+1+2m+1|}{\sqrt{m^2+1}}=2\\ \Rightarrow \frac{|4m+2|}{\sqrt{m^2+1}}=2\\ \Rightarrow |2m+1|=\sqrt{m^2+1}\\ \Rightarrow 4m^2+1+4m=m^2+1\\ \Rightarrow 3m^2+4m=0\\ \Rightarrow m(3m+4)=0\\ m=0,m=\frac{-4}{3}\end{array}$

$\begin{array}{l}Ifm=0from\left(3\right)y-1=0(x+2)\Rightarrow y-1=0\\ Ifm=\frac{-4}{3}from\left(3\right)y-1=\frac{-4}{3}(x+2)\\ \Rightarrow 3y-3=-4x-8\\ \Rightarrow 4x+3y+5=0\end{array}$