Question

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

Answer 1

The correct answer is 624
Shortest distance between the lines
2x+6​​=3y−6​​=4z−6​​
3x−λ​=4y−26​​=52+26​​ is 6
Vector along line of shortest distance
=∣∣​i23​j34​k45​∣∣​, ⇒−i^+2j^​−k (its magnitude is 6​ )
Now 6​1​∣∣​6​+λ23​6​34​−36​45​∣∣​=±6
⇒λ=−26​,106​
So, square of sum of these values is 624 .