JEE Main 2026 — Jan 23, Shift 1 — Mathematics solution

JEE Main2026Jan 23, Shift 1Mathematics

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 $(4, 9, α)$ respectively and C is at a distance of 10 units from B. The area (in sq. units) of $Δ A B C$ is :

A
$513$
B
$1513$
C
$2013$
D
$1013$

Solution

Step 1 — Find $α$ . Rewrite the line as $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ (direction vector $⟨ 1, 2, 3 ⟩$ ). For B $= (4, 9, α)$ on the line: $\frac{4}{1} = \frac{9 - 1}{2} = 4 = \frac{α - 2}{3}$ , so $α = 14$ .

Step 2 — Foot of perpendicular D from A. Let D $= (λ, 2 λ + 1, 3 λ + 2)$ . Then $A D = (λ - 1, 2 λ - 5, 3 λ - 1)$ . Perpendicularity to $⟨ 1, 2, 3 ⟩$ :

$(λ - 1) + 2 (2 λ - 5) + 3 (3 λ - 1) = 0 \Rightarrow 14 λ - 14 = 0 \Rightarrow λ = 1$ .

So D $= (1, 3, 5)$ and $∣ A D ∣ = 0 + 9 + 4 = 13$ .

Step 3 — Area. Base $∣ B C ∣ = 10$ , height $∣ A D ∣ = 13$ :

Area $= \frac{1}{2} \cdot 10 \cdot 13 = 513$ .

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