Exam

JEE Main

Year

2026

Date

23 January

Shift

Morning Shift

Subject

Mathematics

Chapter

Three Dimensional Geometry

JEE Main 2026 · 23 January · Morning Shift · Mathematics · Three Dimensional Geometry

The vertices B and C of a triangle ABC lie on the line...

The vertices B and C of a triangle ABC lie on the line $\frac{x}{1}=\frac{1-y}{-2}=\frac{\mathrm{z}-2}{3}$. 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 :

Step-by-step solution

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

$\overrightarrow{AD} \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) = 0$
$(\lambda - 1)\hat{i} + (2\lambda - 5)\hat{j} + (3\lambda - 1)\hat{k} = \overrightarrow{AD}$
$\Rightarrow \lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0$
$\Rightarrow 14\lambda = 14 \Rightarrow \lambda = 1$
$D = (1, 3, 5)$
$AD = \sqrt{3^2 + 2^2} = \sqrt{13}$
$Ar(\triangle ABC) = \dfrac{1}{2} \times \sqrt{13} \times 10 = 5\sqrt{13}$