The vertices B and C of a triangle ABC lie on the line x/1=1-y/-2=z-2/3 . The coordinates of A and B are (1,6,3) and (4,9, α) respectively and C is at a distance of 10 units from B . The area (in sq. units) of triangle ABC is:

Mathematics · Straight lines

JEE Main 2026 — 23 January, Morning Shift — Question 15

The vertices B and C of a triangle ABC lie on the line $\frac{x}{1}=\frac{1-y}{-2}=\frac{z-2}{3}$ . The coordinates of $A$ and $B$ are $(1,6,3)$ and $(4,9, \alpha)$ respectively and C is at a distance of 10 units from B . The area (in sq. units) of $\triangle \mathrm{ABC}$ is:

Option A:
$5 \sqrt{13}$
Correct
Option B:
$15 \sqrt{13}$
Option C:
$20 \sqrt{13}$
Option D:
$10 \sqrt{13}$

Answer: A

Step-by-step solution

$\frac{4}{1}=\frac{9-1}{2}=\frac{\alpha-2}{3} \Rightarrow \alpha=14$

$\overrightarrow{\mathrm{AD}} \cdot(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=0$

$(\lambda-1) \hat{\mathrm{i}}+(2 \lambda-5) \hat{\mathrm{j}}+(3 \lambda-1) \hat{\mathrm{k}}=\overrightarrow{\mathrm{AD}}$

$\Rightarrow \lambda-1+4 \lambda-10+9 \lambda-3=0$

$\Rightarrow 14 \lambda=14 \Rightarrow \lambda=1$

$\mathrm{D}=(1,3,5)$

$\mathrm{AD}=\sqrt{3^{2}+2^{2}}=\sqrt{13}$

$\operatorname{Ar}(\triangle \mathrm{ABC})=\frac{1}{2} \times \sqrt{13} \times 10=5 \sqrt{13}$

Answer key and solution verified before publishing.

Practise Straight lines

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 2026
Paper: 23 January, Morning Shift
Subject: Mathematics
Chapter: Straight lines
Topic: Special Points in a Triangle

A rectangle is formed by the lines x=0, y=0 , x=3 and y=4 . Let the line L be perpendicular to 3 x+y+6=0 and divide the area of the…

Answer: A

Step-by-step solution

Practise Straight lines

More Straight lines questions from this paper