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

Answer: 384

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
$=\left|\begin{array}{ccc}i& j& k\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|$, $\Rightarrow -\hat{i}+2\hat{j}-k$ (its magnitude is $\sqrt{6}$ )
Now $\frac{1}{\sqrt{6}}\left|\begin{array}{ccc}\sqrt{6}+\lambda & \sqrt{6}& -3\sqrt{6}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|=\pm 6$
$\Rightarrow \lambda =-2\sqrt{6},10\sqrt{6}$
So, square of sum of these values is $384$ .