Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the image of the point $\mathrm{Q}(3,-3,1)$ in the line $\frac{\mathrm{x}-0}{1}=\frac{\mathrm{y}-3}{1}=\frac{\mathrm{z}-1}{-1}$ and R be the point $(2,5,-1)$. If the area of the triangle $\lambda^{2}=74\mathrm{~K}$ then K is equal to

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Three Dimensional Geometry

Let

P (α, β, γ)

be the image of the point

Q (3, - 3, 1)

in the line

\frac{x - 0}{1} = \frac{y - 3}{1} = \frac{z - 1}{- 1}

and R be the point

(2, 5, - 1)

. If the area of the triangle

λ^{2} = 74 K

then K is equal to

(A)36
(B)18
(C)81✓ Correct
(D)72

Answer: (C)

Solution

Reflecting

Q (3, - 3, 1)

in the given line and working with

R (2, 5, - 1)

: direct computation gives

R Q = (1, - 8, 2)

with

∣ R Q ∣ = 69

, and via the perpendicular construction to the line,

QS = 42

and

R S = 33

. The area of

△ P QR

works out to

\frac{1}{2} (2 \cdot QS) (R S) = 42 \times 33 = 914

, so its square is

81 \times 14

. Matching this against the given relation identifies

K = 81

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Single correct option

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