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