If the distance of the point (a,2,5) from the image of the point (1,2,7) in the line 1x=1y−1=2z−2 is 4, then the sum of all possible values of a is equal to:
(A)11
(B)9
(C)6✓ Correct
(D)4
Answer: (C)
Solution
The line has direction (1,1,2) through (0,1,2); the foot of the perpendicular from P(1,2,7) is (λ,1+λ,2+2λ) with (1−λ)+(1−λ)+2(5−2λ)=0, giving λ=2 and foot (2,3,6). The image is Q=2(2,3,6)−(1,2,7)=(3,4,5). The distance from (a,2,5) to Q is (a−3)2+4=4, so (a−3)2=12 and a=3±23. The sum of all possible values of a is 6.
JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option