x+2y-3 + k(4x-y+7) =0
x+2y-3+4xk-ky+7k=0
x(1+4k)+y(2-k)+7k-3=0
Slope of this line = -(1+4k)/(2-k)
Also,this line is parallel to the line 5x+4y-20=0
whose slope is -(5/4)
As parallel lines have same slope,
(1+4k)/(2-k)=5/4
K=2/7
Substituting the value of k
Thw reqd eqn is 15x+12y-7=0