The area of the quadrilateral formed by the lines $2\mathrm{x}+3\mathrm{y}+5=0$ , $3\mathrm{x}-2\mathrm{y}+4=0$ ,   $4x+6y-4=0$and $6\mathrm{x}-4\mathrm{y}+3=0$  is ( in square units )

Rewrite the given lines as 
   $2\mathrm{x}+3\mathrm{y}+5=0$  …… (1) 
   $3\mathrm{x}-2\mathrm{y}+4=0$  …… (2)
   $2\mathrm{x}+3\mathrm{y}-2=0$ …… (3)
   $3\mathrm{x}-2\mathrm{y}+\frac{3}{2}=0$…… (4)
Equations (1) and (3) are parallel lines, equations (2) and (4) are parallel lines
Equations (1) and (2) are perpendicular lines, and equations (2) and (3) are perpendicular lines
The figure formed by the lines is rectangle. 
The area of the rectangle formed is the product of the distances between parallel lines

The distance between two parallel lines $\mathrm{ax}+\mathrm{by}+{\mathrm{c}}_1=0$  and  $\mathrm{ax}+\mathrm{by}+{\mathrm{c}}_2=0$ is  $\frac{\left|{\mathrm{c}}_2-{\mathrm{c}}_1\right|}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}$
The distance between two parallel lines (1) and (3) is  $\frac{\left|5+2\right|}{\sqrt{2^2+3^2}}=\frac{7}{\sqrt{13}}$

The distance between other two parallel lines (2) and (4) is  $\frac{\left|\frac{3}{2}-4\right|}{\sqrt{2^2+3^2}}=\frac{5}{2\sqrt{13}}$

Therefore, the area is $\frac{7}{\sqrt{13}}\cdot \frac{5}{2\sqrt{13}}=\boxed{\frac{35}{26}}$