Three Dimensional Geometry · Question 3

JEE Main 2026 (Online) 6th April Morning ShiftMCQHard📈 Topic in 68% papers+4−1

Let the image of the point P(1, 6, a) in the line L : x1= y-12= z-a+1b, b>0, be ( a3, 0, a+c ). If S( , , ), >0, is the point on L such that the distance of S from the foot of perpendicular from the point P on L is 2 14, then + + is equal to :

Answer

(3) 21

Solution

[figure]

Any point of L : (t, 1+2t, a-1+bt); direction d=(1, 2, b)

Image P' ( a3, 0, a+c ); midpoint M of PP' is the foot of perpendicular and lies on L

M= ( 3+a6, 3, 2a+c2 )

y-coordinate : 1+2t=3 t=1 x=1

3+a6=1 a=3

z-coordinate : a-1+b= 2a+c2 2+b= 6+c2 c=2b-2 … (i)

PP'= ( a3-1, -6, c )=(0, -6, c) d

0-12+bc=0 bc=12 … (ii)

From (i) and (ii) : b(2b-2)=12 b2-b-6=0 (b-3)(b+2)=0

b>0 b=3, c=4

Foot M=(1, 3, 5), d=(1, 2, 3), | d|= 14

MS=2 14=2| d| S=M 2(1, 2, 3)

S=(3, 7, 11) or (-1, -1, -1)

>0 S(3, 7, 11)

+ + =3+7+11=21

Answer: (3)

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