Let in a $\triangle \mathrm{ABC}$, the length of the side AC be 6 , the vertex B be ( $1,2,3$ ) and the vertices $\mathrm{A}, \mathrm{C}$ lie on the line $\frac{\mathrm{x}-6}{3}=\frac{\mathrm{y}-7}{2}=\frac{\mathrm{z}-7}{-2}$. Then the area (in sq. units) of $\triangle \mathrm{ABC}$ is

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Three Dimensional Geometry

Let in a

△ ABC

, the length of the side AC be 6 , the vertex B be (

1, 2, 3

) and the vertices

A, C

lie on the line

\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}

. Then the area (in sq. units) of

△ ABC

Answer: (B)

Since

A

and

C

lie on the line

\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}

(direction

d = (3, 2, - 2)

), the area of

△ A B C

equals

\frac{1}{2} \times A C \times h

, where

h

is the perpendicular distance from

B

to this line.

Let

M = (3 λ + 6, 2 λ + 7, - 2 λ + 7)

be a general point on the line. Then

B M = (3 λ + 5, 2 λ + 5, - 2 λ + 4) .

M

is the foot of the perpendicular from

B

when

B M \cdot d = 0

3 (3 λ + 5) + 2 (2 λ + 5) - 2 (- 2 λ + 4) = 0 \Rightarrow 9 λ + 15 + 4 λ + 10 + 4 λ - 8 = 0 \Rightarrow 17 λ + 17 = 0 \Rightarrow λ = - 1.

Then

M = (3, 5, 9)

and

B M = (2, 3, 6)

, giving

h = ∣ B M ∣ = 4 + 9 + 36 = 49 = 7

Area = \frac{1}{2} \times 6 \times 7 = 21.

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option