JEE Main20265 April 2026Evening ShiftMathematicsThree Dimensional GeometryActual
Let a triangle PQR be such that P and Q lie on the line x+3 8 = y-4 2 = z+1 2 and are at a distance of 6 units from R(1, 2, 3) . If ( , , ) is the centroid of PQR , then + + is equal to :
Options
- A4
- B5
- C6
- D8
Correct answer
C. 6
Step-by-step solution
The equation of the given line is x+3 8 = y-4 2 = z+1 2 . Let x+3 4 = y-4 1 = z+1 1 = . Any point on this line can be taken as S(4 - 3, + 4, - 1) . Since the points P and Q lie on this line and are at a distance of 6 units from R(1, 2, 3) , we have RS^2 = 36 . (4 - 3 - 1)^2 + ( + 4 - 2)^2 + ( - 1 - 3)^2 = 36 (4 - 4)^2 + ( + 2)^2 + ( - 4)^2 = 36 16( ^2 - 2 + 1) + ( ^2 + 4 + 4) + ( ^2 - 8 + 16) = 36 18 ^2 - 36 + 36 = 36 18 ( - 2) = 0 = 0 or = 2 For = 0 , the point is P(-3, 4, -1) . For = 2 , the point is Q(5, 6, 1) .