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 $\triangle ABC$ is:

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Three Dimensional Geometry

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

△ A B C

is:

Answer: (A)

Since B lies on the line

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

, we have

\frac{4}{1} = \frac{1 - 9}{- 2} = \frac{α - 2}{3} = 4

, giving

α = 14

. So

B = (4, 9, 14)

.

The line has direction vector

d = (1, 2, 3)

. Let D be the foot of perpendicular from A to the line.

A point on the line:

(λ, 1 + 2 λ, 2 + 3 λ)

. Vector

A D = (λ - 1, 2 λ - 5, 3 λ - 1)

.

For perpendicularity:

A D \cdot d = 0

gives

(λ - 1) + 2 (2 λ - 5) + 3 (3 λ - 1) = 0

, so

14 λ = 14

λ = 1

.

Thus

D = (1, 3, 5)

and

A D = 0 + 9 + 4 = 13

.

Since BC=10 and AD is the height, Area

= \frac{1}{2} \times 10 \times 13 = 513

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Single correct option