The general equation of a straight line is ax+by+c = 0 where the slope of the line is -a/b.

The line in question is parallel to the line 2x + 3y - 12 = 0. The slope of this line is -2/3. As the lines are parallel so they have the same slopes.

Therefore the line we are looking for is the form of 2x + 3y + c = 0. As this line passes through the point (3,-9), putting these coordinates in the equation of the line will satisfy it.

So we have 2(3) + 3 (-9) + c =0 => 6-27+c=0 => c = 21

So the equation of the line is

2x + 3y + 21 = 0.

The line in question is parallel to the line 2x + 3y - 12 = 0. The slope of this line is -2/3. As the lines are parallel so they have the same slopes.

Therefore the line we are looking for is the form of 2x + 3y + c = 0. As this line passes through the point (3,-9), putting these coordinates in the equation of the line will satisfy it.

So we have 2(3) + 3 (-9) + c =0 => 6-27+c=0 => c = 21

So the equation of the line is

2x + 3y + 21 = 0.