If the length of shortest distance between the two lines $\frac{1}{2}(x-1)=\frac{1}{4}(y-3)=z+2$ and $3x–y–2z+4=0,2x+y+z+1=0$ is s. Then the values of $\sqrt{14}\times s=\_\_\_\_$

Any plane through the second line is $(3x-y-2z+4)+\lambda (2x+y+z+1)=0$ or $(3+2\lambda )x+(\lambda -1)y+(\lambda -2)z+4+\lambda =0.$

If this plane is parallel to the first line then its normal must be at right angle to first line. 

$\Rightarrow (3+2\lambda )2+(\lambda -1)4+(\lambda -2)0\Rightarrow \lambda =0$

$\Rightarrow$ equation of plane through the second line and parallel to the first line is  $3x–y–2z+4=0 ....\mathrm{..}...\left(i\right)$

Now the required shortest distance S = perpendicular distance of a point (1, 3, – 2) on the first line to the plane

$3x–y–2z+4=0=\frac{8}{\sqrt{14}}$