We have $AD\perp OB, \frac{AP}{PD}=\frac{3}{4}$

Equation of line $OB, y=3x$

Equation of line $AD, y+1=-\frac{1}{3}(x+3)$

$\Rightarrow y=\frac{-x}{3}-2$

Intersection point of $AD$ and $OB$ is $D$

$D\left(\frac{-3}{5},\frac{-9}{5}\right)$

Now, $A(-3, -1), D\left(-\frac{3}{5},\frac{-9}{5}\right), AP:PD = 3:4$

So, using section formula

 $P\left(\frac{-{\displaystyle \frac{9}{5}}-12}{7},\frac{{\displaystyle \frac{-27}{5}}-4}{7}\right)\equiv P\left(\frac{-69}{35},\frac{-47}{35}\right)$

So, equation of line $L$ is $y=3x+k$

Using point $P, K=\frac{160}{35}=\frac{32}{7}$

$L\equiv y=3x+\frac{32}{7}$