JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Three Dimensional Geometry

Let

d

be the distance of the point of intersection of the lines

\frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1}

and

\frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2}

from the point

(7, 8, 9)

. Then

d^{2} + 6

is equal to

(A)72
(B)69
(C)75✓ Correct
(D)78

Answer: (C)

Solution

Line 1:

(3 λ - 6, 2 λ, λ - 1)

; line 2:

(4 μ + 7, 3 μ + 9, 2 μ + 4)

. Solving the intersection gives

μ = - 1, λ = 3

, so the point is

(3, 6, 2)

.
Distance from

(7, 8, 9)

d^{2} = (3 - 7)^{2} + (6 - 8)^{2} + (2 - 9)^{2} = 16 + 4 + 49 = 69

. Hence

d^{2} + 6 = 75.

JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Single correct option

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