If the family of straight lines  $ax+by+c=0$, where $2a+3b=4c$  is concurrent at the point $P(l,m)$  , then the foot of the perpendicular drawn from  $P$  to the line  $x+y+1=0$  is

We have, $ax+by+c=0$  and  $2a+3b=4c$
Putting  $c=\frac{2a+3b}{4}$  in  $ax+by+c=0$ , we get
 $$\begin{gathered} ax+by+\frac{2a+3b}{4}=0 \\ \Rightarrow 4ax+4by+2a+3b=0 \\ \Rightarrow a(4x+2)+b(4y+3)=0 \\ \Rightarrow (4x+2)+\frac{b}{a}(4y+3)=0 \end{gathered}$$
This is of the form $L_1+\lambda L_2=0$ 
So, set of lines are 
$4x+2=0$  and  $4y+3=0$
These two lines intersect at  $(\frac{-1}{2},\frac{-3}{4})$
 $\therefore P(l,m)=(\frac{-1}{2},\frac{-3}{4})$
 
Equation of line passing through  $(\frac{-1}{2},\frac{-3}{4})$
and perpendicular to $x+y+1=0$  is
  $y+\frac{3}{4}=1(x+\frac{1}{2})$
 $\Rightarrow 4y=4x-1\Rightarrow 4x-4y-1=0$
On solving  $x+y+1=0$
and  $4x-4y-1=0,$
we get  $Q(-\frac{3}{8},-\frac{5}{8})$