We have, $a x+b y+2=0$ and $a x+b y+5=0$ are parallel lines.
$$
a x+b y+2=0 \Rightarrow y=\frac{-a}{b} x-\frac{2}{b}
$$
and $a x+b y+5=0 \Rightarrow y=\frac{-a}{b} x \frac{-5}{b}$
So, $m_1=\frac{-a}{b}, c_1=\frac{-2}{b}, c_2=-\frac{5}{b}$
Now, again
$$
\begin{aligned}
& \qquad x+d y+3=0 \Rightarrow y=\frac{-c}{d} x-\frac{3}{d} \\
& \text { and } c x+d y+7=0 \Rightarrow y=\frac{-c}{d} x-\frac{7}{d}
\end{aligned}
$$
So, $m_2=\frac{-c}{d}, d_1=\frac{-3}{d}, d_2=\frac{-7}{d}$
Then, area of parallelogram $=\left|\frac{\left(c_1-c_2\right)\left(d_1-d_2\right)}{m_1-m_2}\right|$
$$
\begin{aligned}
& =\left|\frac{\left(\frac{-2}{b}+\frac{5}{b}\right)\left(\frac{-3}{d}+\frac{7}{d}\right)}{-\frac{a}{b}+\frac{c}{d}}\right| \\
& =\left|\frac{\frac{3}{b} \times \frac{4}{d}}{\frac{-(a d-b c)}{b d}}\right|=\left|\frac{-12}{a d-b c}\right|=\frac{12}{|a d-b c|} .
\end{aligned}
$$