Consider the lines $\mathrm{L}_{1}: \mathrm{x}-1=\mathrm{y}-2=\mathrm{z}$ and $\mathrm{L}_{2}: \mathrm{x}-2=\mathrm{y}=\mathrm{z}-1$. Let the feet of the perpendiculars from the point $\mathrm{P}(5,1,-3)$ on the lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ be Q and R respectively. If the area of the triangle PQR is A , then $4 \mathrm{~A}^{2}$ is equal to :

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Three Dimensional Geometry

Consider the lines

L_{1} : x - 1 = y - 2 = z

and

L_{2} : x - 2 = y = z - 1

. Let the feet of the perpendiculars from the point

P (5, 1, - 3)

on the lines

L_{1}

and

L_{2}

be Q and R respectively. If the area of the triangle PQR is A , then

4 A^{2}

is equal to :

(A)139
(B)147✓ Correct
(C)151
(D)143

Answer: (B)

Solution

Both lines have direction

(1, 1, 1)

, with

L_{1}

through

(1, 2, 0)

and

L_{2}

through

(2, 0, 1)

. For the foot

Q

L_{1}

, take

(1 + λ, 2 + λ, λ)

and set

P Q \cdot (1, 1, 1) = 0

, giving

λ = 0

, so

Q = (1, 2, 0)

. Similarly

R = (2, 0, 1)

. Then

P Q = (- 4, 1, 3)

and

P R = (- 3, - 1, 4)

, whose cross product is

(7, 7, 7)

with magnitude

73

. The area is

A = \frac{1}{2} \cdot 73

, so

4 A^{2} = 4 \cdot \frac{147}{4} = 147

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Single correct option

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