Question 1132638
The slope of the line is -12/5, and the slope of the line perpendicular to that is 5/12.

{{{graph(300,300,-10,10,-10,10,-2.4x+2,5x/12)}}}

this is the general form of the line perpendicular to the given line.
We want to go 3 units out on the perpendicular.  
Suppose the perpendicular is to the line at the x-intercept of (5/6, 0). Then the line can be shown using the point slope formula of y-y1=m(x-x1), where m is slope and (x1, y1) are (5/6, 0) to be y=(5/12)x-(25/72)
{{{graph(300,300,-2,2,-2,2,-2.4x+2,5x/12-(25/72))}}}

Going to the right 3 units is a right triangle with the line 3 units away.  The length of the hypotenuse is where the x-intercept is for that line.  The angle for a slope of 5/12 is where the tangent of an angle is 5/12, which is arc tan (5/12) or 22.62 degrees.

So cos 22.62 =3/hypot
hypot=3/cos 22.62=3.25 units away from (5/6, 0) so the x-intercept of that line is 3.25+5/6 or (4.08, 0)
We know the slope of this parallel line is -2.4, and we have a point, so the equation of the line is y=-2.4(x-4.08)=y=-2.4x+9.08.  

{{{graph(300,300,-10,10,-10,10,-2.4x+2,-2.4x+9.08)}}}

The distance to the x-intercept in the other direction is -3.25 units, so the x-intercept is at -3.25+5/6=-2.42 or (-2.42, 0)

The equation of this line is y=-2.4(x+2.42)=-2.4x-5.8

{{{graph(300,300,-10,10,-10,10,-2.4x+2,-2.4x+9.08,-2.4x-5.8)}}}

The distance between the two y-intercepts is 9.08-(-5.8)=14.88 units ANSWER