Find the vector perpendicular to the line 3x-5y+2=0.
Write in the form y = mx + c, i.e. y = (3/5)x + 2/5, which has gradient = 3/5.
The gradients of perpendicular lines have a product of -1. So the gradient of your perpendicular is -5/3.
So any line in the family of lines y = (-5/3)x + c will be perpendicular. For simplicity we'll just consider y = (-5/3)x.
If you note from the slope that this perpendicular line moves down 5 units as you move to the right 3 units, then you can write this line in its vector form t(3i - 5j), where t is an arbitrary constant. If you know a point that this line goes through, say (x_0, y_0), then the line can be written as (x_0 + 3t)i + (y_0 - 5t)j.