The point on $Y-$axis whose perpendicular distance from the line  $3x+4y+1=0$  is 5 units

Perpendicular distance from point on $Y-$axis to the line $3x+4y+1=0$  is 5 units

$\left|\frac{3\left(0\right)+4\left(y\right)+1}{\sqrt{3^2+4^2}}\right|=5$

$4y+1=5\times 5$  (or)  $4y+1=-5\times 5$

$4y=24\to y=6$  (or)  $4y=-26\to y=\frac{-26}{4}=\frac{13}{2}$

Point on $Y-$axis is $(0,y)=(0,6)$  (or)  $(0,\frac{-13}{2})$.