Find the maximum possible number of points of intersection of perpendicular lines I know many such questions have already been asked but this one seemed to be a very messy and a lengthy question for me to solve. The question is as follows 

Consider seven different points $P_1,P_2,P_3,P_4,Q_1,Q_2,Q_3$ in plane such that $P_1,P_2,P_3,P_4$ are on straight line $'l'$; and $Q_1,Q_2,Q_3$ are non collinear points and none of them lies on straight line $'l'$( $Q_1,Q_2,Q_3$ are on same side of line $'l'$)
From each of the three points $Q_1,Q_2,Q_3$ perpendicular lines are drawn to the straight lines formed by joining any two of the given points (excluding the point from which the perpendicular line is drawn). Find the maximum possible number of points of intersection of perpendicular lines ( excluding the points $Q_1,Q_2,Q_3$ ).

After a lot of messy work and a turmoil of 3 hours I am getting the answer as $283$. I pretty much think that the answer might be correct.  But I wanted to know thoughts of members on this site over how to approach this problem without expending so much of time because such question was asked in our examination papers a few years back and I cannot afford so much time on a single question in exam. So please share your thoughts over this question 
 A: INTERSECTION POINTS BETWEEN 2 PERPENDICULAR LINES DRAWN FROM $Q_1,Q_2,Q_3$
Calculating no of perpendicular lines for each $Q_1,Q_2$ and $Q_3$ point 
CAT $1$ -  line  made by joining $P$ points
CAT $2$- lines made by joining $1$ point from $P$ and $1$ point from $Q$
CAT $3$- lines made by joining points from $Q.$
For $Q_1$ point
CAT $1$ - since all the $P$ points lie in same line $'l'$ there is only $1$ single line and there will be only $1$  perpendicular drawn from $Q_1$.
CAT 2 - Total no. of perpendicular drawn from $Q_1$ is given by $^4C_1 * ^2C_1 =8$ ways. (selecting $1$ point from $P_1, P_2 ,P_3 , P_4$ and another point from $Q_2 ,Q_3$)
CAT $3$ – only $1$ perpendicular will be drawn to the line $Q_2Q_3 $
Total no. of perpendiculars drawn from $Q_1=1+8+1=10$
Total no of perpendiculars lines$ = 10*3 = 30$
No of intersection points from $30$ perpendicular lines = $^{30}C_2=435.$  But this is not true
Case $1$- some of these lines are parallel to each other therefore they will not intersect
(a) Lines from $Q_1 ,Q_2 ,Q_3$ to line $‘l'$.Total points to be cut$= ^3C_2=3$
(b) Lets connect a line between $Q1$ and $P_1$ the perpendicular are drawn  from $Q_2$ and $Q_3$ to it will be parallel. such lines can be given by $^3C_1*^4C_1=12$
Case $2$: From each of the $3$ points ($Q_1,Q_2,Q_3$), $10 $ perpendicular lines are drawn through it. So it is the point of intersection of $10$ perpendicular lines. We have counted this point $^{10}C_2 =45$ times as a point of intersection of perpendicular lines. So the number of points of intersection should be cut by $3×45=135$(as it is mentioned that $Q_1,Q_2,Q_3$ have to be excluded)
Case $3$: The orthocenter of triangle formed by $Q_1Q_2Q_3 $has been counted $3$ times instead of $1$
Total number of intersection $= 435 - 3 - 12 - 135 - 2 = 283$
A: INTERSECTION POINT OF PERPENDICULAR LINE FROM Q1,Q2,Q3 TO THE LINE 
first we will calculate how many lines are made using the above points 
it comes from three categories:-
CAT 1- lines made by joining P point.
CAT 2- lines made by joining 1 point from P and 1 point from Q
CAT 3- lines made by joining points from Q.
CAT 1 - since all the P points lie in same line 'l' there is only 1 single line and the perpendiculars drawn from Q1,Q2,Q3 intersect in exactly 3 places.
CAT 2 - total no. of lines is given by 3C1 * 4C1 =12 ways. Perpendiculars are drawn from the other 2 points giving the total no. of intersection points= 12 * 2= 24
CAT 3 - total no. of lines is given by 3C2 = 3 ways. Perpendicular is drawn from the remaining point. Hence no of intersection = 3
total no of points = 3 + 24 + 3 = 30 points
