QJEEM20Mad9a8: JEE Main 2020 Mathematics - Class 12 JEEM20Mad9a8 Solution: If (a, b, c) is the image of the point (1, 2, -3) in the line , …

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