The area of the triangle formed by the lines 2x +y -4=0, 3x+2y-5=0, x+y=1=0 is

$\begin{array}{l}Givenlines2x+y-4=0--\left(1\right),3x+2y-5=0--\left(2\right),x+y+1=0--\left(3\right)\\ Solving\left(1\right)\&\left(2\right)\\ x y 1\\ \begin{array}{cccc}1& -4& 2& 1\\ 2& -5& 3& 2\end{array}\\ \Rightarrow \frac{x}{-5+8}=\frac{y}{-12+10}=\frac{1}{4-3}\\ \Rightarrow \frac{x}{3}=\frac{y}{-2}=\frac{1}{1}\\ \therefore (x,y)=(3,-2)\\ Solving\left(2\right)\&\left(3\right)\\ x y 1\\ \begin{array}{cccc}2& -5& 3& 2\\ 1& 1& 1& 1\end{array}\\ \Rightarrow \frac{x}{2+5}=\frac{y}{-5-3}=\frac{1}{3-2}\\ \Rightarrow \frac{x}{7}=\frac{y}{-8}=\frac{1}{1}\\ \therefore (x,y)=(7,-8)\\ Solving\left(1\right)\&\left(3\right)\end{array}$

$\begin{array}{l}xy1\\ \begin{array}{cccc}1& -4& 2& 1\\ 1& 1& 1& 1\end{array}\\ \Rightarrow \frac{x}{1+4}=\frac{y}{-4-2}=\frac{1}{2-1}\\ \Rightarrow \frac{x}{5}=\frac{y}{-6}=\frac{1}{1}\\ \therefore (x,y)=(5,-6)\\ A(3,-2),B(7,-8),C(5,-6)\\ Areaof{\Delta }^{le}ABc=\frac{1}{2}\left|\begin{array}{cc}{x}_1-x_2& x_1-x_3\\ y_1-y_2& y_1-y_3\end{array}\right|\\ \Rightarrow \frac{1}{2}\left|\begin{array}{cc}3-7& 3-5\\ -2+8& -2+6\end{array}\right|\\ \Rightarrow \frac{1}{2}\left|\begin{array}{cc}-4& -2\\ 6& 4\end{array}\right|\\ \Rightarrow \frac{1}{2}|-16+12|\Rightarrow \frac{4}{2}=2\end{array}$