The shortest distance between the lines ${{x + 2} \over 1} = {y \over { - 2}} = {{z - 5} \over 2}$ and ${{x - 4} \over 1} = {{y - 1} \over 2} = {{z + 3} \over 0}$ is :

Given, the lines are <br/><br/>$$ \begin{aligned} \frac{x+2}{1} & =\frac{y}{-2}=\frac{z-5}{2} ~~~~..........(i)\\\\ \text { and } \frac{x-4}{1} & =\frac{y-1}{2}=\frac{z+3}{0} ~~~~..........(ii) \end{aligned} $$ <br/><br/>Formula for shortest distance between two skew-lines, <br/><br/>$$ \begin{aligned} S D & =\left|\frac{\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right|}{\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right|}\right|=\left|\frac{\left|\begin{array}{ccc} 6 & 1 & -8 \\ 1 & -2 & 2 \\ 1 & 2 & 0 \end{array}\right|}{\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -2 & 2 \\ 1 & 2 & 0 \end{array}\right|}\right| \\\\ & =\left|\frac{6(-4)-1(-2)-8(4)}{|-4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}|}\right| \\\\ & =\left|\frac{-24+2-32}{\sqrt{36}}\right| \\\\ & =\left|\frac{-54}{6}\right|=|-9|=9 \end{aligned} $$