Let P be the image of the point Q(7, -2, 5) in the line L: x-1/2 = y+1/3 = z/4 and R(5, p, q) be a point on L. Then the square of the area of triangle PQR is \ \ \ \ \ .

Mathematics · 3D Geometry

JEE Main 2025 — 24 January, Evening Shift — Question 20

Let P be the image of the point Q(7, -2, 5) in the line $L: \frac{x-1}{2} = \frac{y+1}{3} = \frac{z}{4}$ and R(5, p, q) be a point on L. Then the square of the area of $\triangle$ PQR is $\_\_\_\_\_$ .

Answer: 957

Numerical answer — enter this value.

Step-by-step solution

\text{Let\; } R\; (2\lambda + 1, 3\lambda - 1, 4\lambda)

2\lambda + 1 = 5

\lambda = 2

R(5, 5, 8)

\text{Let\; } T(2\lambda + 1, 3\lambda - 1, 4\lambda)

\overrightarrow{QT} = (2\lambda - 6) \hat{i} + (3\lambda + 1) \hat{j} + (4\lambda - 5) \hat{k}

\vec{b} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}

\overrightarrow{QT} \cdot \vec{b} = 0

4\lambda - 12 + 9\lambda + 3 + 16\lambda - 20 = 0

$\lambda=1$

$\mathrm{T}(3,2,4)$

$\mathrm{QT}=\sqrt{33} \quad \mathrm{RT}=\sqrt{29}$

$(\text { area\; of \;} \triangle \mathrm{PQR})^{2}=\left(\frac{1}{2} \sqrt{29} .2 \sqrt{33}\right)^{2}$

$=957$

Answer key and solution verified before publishing.

Practise 3D Geometry

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 24 January, Evening Shift
Subject: Mathematics
Chapter: 3D Geometry
Topic: Intersection of lines, line & plane.