Official solution
Equation of line : \[{{x + 1} \over 2} = {{y - 3} \over { - 2}} = {z \over { - 1}}\] = \[\lambda \] (Assume)
A point on line L is
= R(2\[\lambda \] -- 1, --2\[\lambda \] + 3, --\[\lambda \])
DR's of PR = < 2\[\lambda \] -- 2, --2\[\lambda \] + 1, --\[\lambda \] + 3 >
\[ \because \] PR is perpendicular to line L
\[ \therefore \] 2(2\[\lambda \] --2) --2 (--2\[\lambda \] + 1) --1 (--\[\lambda \] + 3) = 0
\[ \Rightarrow \] 4\[\lambda \] -- 4 + 4\[\lambda \] -- 2 + \[\lambda \] -- 3 = 0
\[ \Rightarrow \] 9\[\lambda \] -- 9 = 0
\[ \Rightarrow \] \[\lambda \] = 1
\[ \therefore \] Coordinate of foot of perpendicular = R = (1, 1, --1)
As R is the midpoint of line PQ, so
\[{{a + 1} \over 2} = 1\] \[ \Rightarrow \] a = 1
\[{{b + 2} \over 2} = 1\] \[ \Rightarrow \] b = 0
\[{{c - 3} \over 2} = - 1\] \[ \Rightarrow \] c = 1
\[ \therefore \] a + b + c = 2