Given:

$\left\{\begin{array}{l}ax+by+2=0 ...\left(\mathrm{i}\right)\\ ax+by+5=0 ....\left(\mathrm{ii}\right)\end{array}\to \text{Parallel lines}\right.$

$\left\{\begin{array}{l}cx+dy+3=0 ....\left(\mathrm{iii}\right)\\ cx+dy+7=0 ....\left(\mathrm{iv}\right)\end{array}\to \text{Parallel lines}\right.$

Opposite sides are parallel in parallelogram.

Let $d_1$ be the distance between $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right)$, then

$d_1=\frac{5-2}{\sqrt{a^2+b^2}}=\frac{3}{\sqrt{a^2+b^2}}$

Let $d_2$ be the distance between $\left(\mathrm{iii}\right) \& \left(\mathrm{iv}\right)$, then

$d_2=\frac{7-3}{\sqrt{c^2+d^2}}=\frac{4}{\sqrt{c^2+d^2}}$

Area of parallelogram$=\left(\frac{3}{\sqrt{a^2+b^2}}\right)\times \left(\frac{4}{\sqrt{c^2+d^2}}\right)$

                                   $=\frac{12}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}}$