The shortest distance (in units) between the lines $\frac{1-x}{1}=\frac{2y-10}{2}=\frac{z+1}{1}$ and $\frac{x-3}{-1}=\frac{y-5}{1}=\frac{z-0}{1}$ is:

$\sqrt{\frac{14}{3}}$

The correct answer is Option (4) → $\sqrt{\frac{14}{3}}$

The given line are,

$\frac{1-x}{1}=\frac{2y-10}{2}=\frac{z+1}{1}$

$⇒x=1-t,y=\frac{10+2t}{2},z=-1+t$

∴ Direction vector, $d_1=(-1,1,1)$

for second line,

$\frac{x-3}{-1}=\frac{y-5}{1}=\frac{z-0}{1}$

$⇒x=3-s,y=5+s,z=s$

$∴d_2=(-1,1,1)$

Point on 1st line, $r_1=(1,5,-1)$

Point on 2nd line, $r_2=(3,5,0)$

$d=\frac{|(r_2-r_1)×d_1|}{|d_1|}$

$=\frac{|(2,0,1)×(-1,1,1)|}{\sqrt{(-1)^2+(1)^2+(1)^2}}$   $\begin{vmatrix}\hat i&\hat j&\hat k\\2&0&1\\-1&1&1\end{vmatrix}$

$=\frac{|-1\hat i-3\hat j+2\hat k|}{\sqrt{3}}$

$=\frac{\sqrt{(-1)^2+(-3)^2+(2)^2}}{\sqrt{3}}$

$=\sqrt{\frac{14}{3}}$