If a straight line perpendicular to $2x-3y+7=0$ forms a triangle with the coordinate axes whose area is 3 sq. units then the equation of the lines are 

Let Equation of line perpendicular to $2x-3y+7=0$ is $3x+2y+k=0$

the area formed by the line $3x+2y+k=0$ with coordinate axes is $\frac{k^2}{2\left|3\times 2\right|}$

$$\begin{gathered} \Rightarrow \frac{k^2}{2\left|3\times 2\right|}=3 \\ \Rightarrow k^2=36 \\ \Rightarrow k=\pm 6 \\ required line is 3x+2y=\pm 6 \end{gathered}$$