If the shortest distance between the lines $\frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$ is 1, then the sum of all possible values of $\lambda$ is :

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Three Dimensional Geometry

If the shortest distance between the lines

\frac{x - λ}{- 2} = \frac{y - 2}{1} = \frac{z - 1}{1}

and

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

is 1, then the sum of all possible values of

λ

is :

Answer: (B)

Line 1 passes through

(λ, 2, 1)

with direction

(- 2, 1, 1)

; Line 2 passes through

(3, 1, 2)

with direction

(1, - 2, 1)

.
Vector between the given points:

(3 - λ, - 1, 1)

. Cross product of the directions:

(- 2, 1, 1) \times (1, - 2, 1) = (1 (1) - 1 (- 2), 1 (1) - (- 2) (1), (- 2) (- 2) - 1 (1)) = (3, 3, 3)

Shortest distance

= \frac{∣ ( 3 - λ , - 1 , 1 ) \cdot ( 3 , 3 , 3 ) ∣}{∣ ( 3 , 3 , 3 ) ∣} = \frac{∣3 ( 3 - λ ) - 3 + 3∣}{3 3} = \frac{∣ 3 - λ ∣}{3}

.
Setting this equal to 1:

∣ 3 - λ ∣ = 3 ⟹ λ = 0

λ = 23

. Sum of all possible values:

0 + 23 = 23

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Single correct option