If the shortest between the lines $\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$ and $\frac{x-\lambda}{3}=\frac{y-2 \sqrt{6}}{4}=\frac{z+2 \sqrt{6}}{5}$ is 6, then the square of sum of all possible values of $\lambda$ is

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Three Dimensional Geometry

If the shortest between the lines

\frac{x + 6}{2} = \frac{y - 6}{3} = \frac{z - 6}{4}

and

\frac{x - λ}{3} = \frac{y - 2 6}{4} = \frac{z + 2 6}{5}

is 6, then the square of sum of all possible values of

λ

Answer: 384

Solution

The lines have directions

b_{1} = (2, 3, 4)

and

b_{2} = (3, 4, 5)

, with

b_{1} \times b_{2} = (- 1, 2, - 1)

and

∣ b_{1} \times b_{2} ∣ = 6

. With base points

(- 6, 6, 6)

and

(λ, 26, - 26)

, their difference is

(λ + 6, 6, - 36)

. The shortest distance is

\frac{∣ ( λ + 6 ) ( - 1 ) + 6 \cdot 2 + ( - 3 6 ) ( - 1 ) ∣}{6} = \frac{∣4 6 - λ ∣}{6} = 6

, so

∣4 6 - λ ∣ = 66

. This gives

λ = - 26

λ = 106

, whose sum is

86

. The square of the sum is

(86)^{2} = 384

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Integer answer

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