Lines are drawn parallel to the line 4x-3y+2=0, at a distance 3/5 from the origin. Then which one of the following points lies on any of these lines?

The correct option is A (-14,23)
Required line is 4x-3y+λ=0
|λ|/5 = 3/5 => λ = ±3
So required equation of line is 4x-3y+3=0 and 4x-3y-3=0
(1) 4(-14)-3(23)+3 = -56 -69 +3 = -122 ≠ 0
(2) 4(-14)-3(23)-3 = -56-69-3 = -128 ≠ 0
Let's check the distance of points from the line 4x - 3y = 0
For (14,13): |4(14) - 3(13)|/5 = |56-39|/5 = 17/5 ≠ 3/5
For (-14, 23): |4(-14) -3(23)|/5 = |-56 -69|/5 = 125/5 = 25 ≠ 3/5
For (-14, -23): |4(-14) - 3(-23)|/5 = |-56 + 69|/5 = 13/5 ≠ 3/5
For (14, -13): |4(14) - 3(-13)|/5 = |56 + 39|/5 = 95/5 = 19 ≠ 3/5
Let's use the equation 4x - 3y + 3 = 0
4(-14) -3(23) + 3 = -56 - 69 + 3 = -122
4x - 3y -3 = 0
4(-14) - 3(23) -3 = -56 -69 -3 = -128
There must be a mistake in the question or options provided.