Let Q and R be two points on the line $\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}$ at a distance $\sqrt{26}$ from the point $P(4,2,7)$. Then the square of the area of the triangle PQR is ______.

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Three Dimensional Geometry

Let Q and R be two points on the line

\frac{x + 1}{2} = \frac{y + 2}{3} = \frac{z - 1}{2}

at a distance

26

from the point

P (4, 2, 7)

. Then the square of the area of the triangle PQR is ______.

Answer: 153

Key concept: Distance from point to line and area of triangle using cross product.

Step 1: Parametric form of line:

(2 λ - 1, 3 λ - 2, 2 λ + 1)

Distance from

P (4, 2, 7)

(2 λ - 5)^{2} + (3 λ - 4)^{2} + (2 λ - 6)^{2} = 26

17 λ^{2} - 68 λ + 51 = 0

λ = 1, 3

Q (1, 1, 3)

and

R (5, 7, 7)

.

Step 2: Find area using vectors.

P Q = (- 3, - 1, - 4), P R = (1, 5, 0)

P Q \times P R = \hat{i} - 3 1 \hat{j} - 1 5 \hat{k} - 4 0 = (20, - 4, - 14)

∣ P Q \times P R ∣ = 400 + 16 + 196 = 612 = 2153

Area =

\frac{1}{2} \cdot 2153 = 153

Square of area =

153

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Integer answer