Solution:

For 3x+2y+4=0

$slope m_{1} = \frac{-a}{b} = \frac{-3}{2}$

For 2x-3y-7=0,

$slope m_{2} = \frac{-a}{b} = \frac{-2}{-3} = \frac{2}{3}$ \begin{aligned} \therefore \tan \theta &=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \ &=\left|\frac{\frac{-3}{2}-\frac{2}{3}}{1+\left(\frac{-3}{2}\right)\left(\frac{2}{3}\right)}\right| \end{aligned}

$\therefore \tan \theta=\infty$

$\therefore \theta=\tan ^{-1}(\infty)$

$\therefore \theta=90^{\circ} \quad or \quad \frac{\pi}{2}$