Let L be the straight line joining the points P(1,2, -1) and Q(2,3,1). Let S be the foot of the perpendicular drawn from the point $\mathrm{R}(4,-1,5)$ to the line L. Another line passing through R intersects L at a point T such that the point S divides the line segment PT internally in the ratio $|\mathrm{PS}|:|\mathrm{ST}|=1: 2$, where $|\mathrm{PS}|$ and $|\mathrm{ST}|$ are the lengths of the line segments PS and ST, respectively. Then which of the following statments is (are) TRUE ? (A) The orthocentre of the triangle PRT is $\left(\frac{23}{5},-4, \frac{31}{5}\right)$ (B) The orthocentre of the traiangle PRT is (4,3,5) (C) The area of the triangle PRT is $6 \sqrt{5}$ (D) The area of the triangle PRT is $18 \sqrt{5}$

(A) The orthocentre of the triangle PRT is $\left(\frac{23}{5},-4, \frac{31}{5}\right)$; (D) The area of the triangle PRT is $18 \sqrt{5}$

(C) The area of the triangle PRT is $6 \sqrt{5}$

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Three Dimensional Geometry

WhatsApp Email

Let L be the straight line joining the points P(1,2, -1) and Q(2,3,1). Let S be the foot of the perpendicular drawn from the point

R (4, - 1, 5)

to the line L. Another line passing through R intersects L at a point T such that the point S divides the line segment PT internally in the ratio

∣ PS ∣ : ∣ ST ∣ = 1 : 2

, where

∣ PS ∣

and

∣ ST ∣

are the lengths of the line segments PS and ST, respectively. Then which of the following statments is (are) TRUE ?
(A) The orthocentre of the triangle PRT is

(\frac{23}{5}, - 4, \frac{31}{5})

(B) The orthocentre of the traiangle PRT is (4,3,5)
(C) The area of the triangle PRT is

65

(D) The area of the triangle PRT is

185

(A)The orthocentre of the triangle PRT is $(\frac{23}{5}, - 4, \frac{31}{5})$ ✓ Correct
(B)The orthocentre of the traiangle PRT is (4,3,5)
(C)The area of the triangle PRT is $65$
(D)The area of the triangle PRT is $185$ ✓ Correct

Answer: (A), (D)

Solution

Line L:

\frac{x - 1}{1} = \frac{y - 2}{1} = \frac{z + 1}{2} = λ

. Point on L:

(1 + λ, 2 + λ, - 1 + 2 λ)

. For foot S,

R S ⊥ (\hat{i} + \hat{j} + 2 \hat{k})

(λ - 3) + (λ + 3) + 2 (2 λ - 6) = 0 \Rightarrow λ = 2

. Thus

S (3, 4, 3)

. Since S divides PT in ratio 1:2, using section formula with

P (1, 2, - 1)

and

T

on L:

T (7, 8, 11)

. Area of

△ P R T = \frac{1}{2} ∣ R P \times R T ∣

where

R P = (- 3, 3, - 6)

and

R T = (3, 9, 6)

. Cross product gives

(72, - 36)

magnitude

365

, so area

= 185

(option D). Since

R S ⊥ P T

, orthocentre H lies on RS. Using

P H ⊥ R T

, we find

H (\frac{23}{5}, - 4, \frac{31}{5})

(option A).

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · One or more correct options

Practise more: all JEE Advanced PYQs on Three Dimensional Geometry →

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Three Dimensional Geometry

Solution

More JEE Advanced questions on Three Dimensional Geometry