find equation of straight line through point A(-2,-7) and having intercepts of length 3 between straight lines 4x+3y=12 and 4x+3y=3

4x + 3y = 12 and 4x + 3y = 3 are parallel lines

Let us first find the distance between the parallel lines

$\frac{3\times 4-3}{\sqrt{4^2+3^2}}=\frac{9}{5}=1.8$

Let A be the angle made by the line with 4x + 3y = 3

$\tan A=\frac{\sqrt{3^2-1.8^2}}{1.8}=\frac{4}{3}$
Let m be the slope of the line

$\tan A=\frac{m-{\displaystyle \frac{4}{3}}}{1+{\displaystyle \frac{4m}{3}}}=\frac{4}{3}\Rightarrow 9m-12=12 + 16m\Rightarrow m=-\frac{24}{7}$
Equation of the line:

$y-\left(-7\right)=-\frac{24}{7}\left(x-\left(-2\right)\right)$
7y + 24x + 97=0