The shortest distance between the lines $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$ lies in the interval

(2, 3]

Given lines pass through the points having position vectors $\vec{a_1} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k}$ and $\vec{a_2} =-2\hat{i} + 4\hat{j} + 5\hat{k}$ and are parallel to vectors $\vec{b_1} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b_2}= -\hat{i} +8\hat{j}+4\hat{k}.$

$∴ \vec{b_1}× \vec{b_2}= \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\2 & 2 & 1\\-1 & 8 & 4\end{vmatrix}=0\hat{i} -9\hat{j} +18\hat{k}$

and $\vec{a_2}-\vec{a_1}= -2\hat{i}+4\hat{j} + 5\hat{k}$

$S. D.=\begin{vmatrix} \frac{(\vec{b_1}×\vec{b_2}).(\vec{a_2}-\vec{a_1})}{|\vec{b_1}×\vec{b_2}|}\end{vmatrix}= \begin{vmatrix} \frac{-2×0-36+90}{\sqrt{0+81+324}}\end{vmatrix}= \frac{504}{\sqrt{405}}=\frac{6}{\sqrt{5}}$