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

Shortest distance between the lines
$\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$
$\frac{x-\lambda}{3}=\frac{y-2 \sqrt{6}}{4}=\frac{2+2 \sqrt{6}}{5}$ is 6
Vector along line of shortest distance
$= \begin{vmatrix} i & j & k \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{vmatrix}$, $\Rightarrow-\hat{ i }+2 \hat{ j }- k$ (its magnitude is $\sqrt{6}$ )
Now $\frac{1}{\sqrt{6}} \begin{vmatrix}\sqrt{6}+\lambda & \sqrt{6} & -3 \sqrt{6} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{vmatrix}=\pm 6$
$\Rightarrow \lambda=-2 \sqrt{6}, 10 \sqrt{6}$
So, square of sum of these values is $384$ .