JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Three Dimensional Geometry

If the lines

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

and

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

intersect at the point P, then the distance of the point P from the plane

z = a

is:

Answer: (B)

Solution

Write a general point of the first line as

(λ + 1, 2 λ + 2, λ - 3)

and of the second as

(2 μ + a, 3 μ - 2, μ + 3)

. Equating the

z

-coordinates gives

λ = μ + 6

, and equating the

y

-coordinates gives

2 λ = 3 μ - 4

. Substituting the first into the second,

2 (μ + 6) = 3 μ - 4

, so

μ = 16

and

λ = 22

. The point of intersection is therefore

P = (23, 46, 19)

, and matching the

x

-coordinates gives

23 = 2 (16) + a

, so

a = - 9

. The distance of

P

from the plane

z = a = - 9

∣19 - (- 9) ∣ = 28

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Single correct option