Given equation of line is:  

       (x/4) +(y/6) = 1

=> 3x + 2y = 12 

=> 2y = -3x + 12

=> y = (-3/2)x + 12

Let the slope of the required line is m.

Again given lines are perpendicular.

So  m*(-3/2) = -1    (Product of slope = -1)

=> m = 2/3

So slope of the line is 2/3.

Again let the point where the lines meets on y-axis is (0, y). So put this point on the line we get

3*0 + 2y = 12

=> 2y = 12

=> y = 12/2

=> y = 6

So the points is (0,6) 

Now the equation of line pasiing through point(0,6) and slope is 2/3 is

      y-6 = (2/3)(x-0)

=> 3(y-6) = 2x

=> 3y -18 = 2x

=> 2x - 3y +  18 = 0

So the equation of line is: 2x - 3y +  18 = 0