The shortest distance between the lines \(1+x=2 y=-12 z\) and
\(x=y+2=6 z-6\) is

Shortest distance between the lines
\(
\begin{array}{c}
\frac{x+1}{1}=\frac{y}{\left(\frac{1}{2}\right)}=\frac{z}{\left(\frac{-1}{12}\right)} \text { and } \frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\left(\frac{1}{6}\right)} \text { is } \\
\begin{array}{c}
0+1-2-0 \\
1 \\
\frac{1}{2} & \frac{-1}{12} \\
1 & \frac{1}{6}
\end{array} \mid \\
d=\frac{\sqrt{\left(\frac{1}{6}+\frac{1}{12}\right)^{2}+\left(1-\frac{1}{2}\right)^{2}+\left(\frac{1}{12}+\frac{1}{12}\right)^{2}}}{12}
\end{array}
\)
\(=\frac{\left|\begin{array}{ccc}1 & -2 & 1 \\ 1 & \frac{1}{2} & \frac{-1}{12} \\ 1 & 1 & \frac{1}{6}\end{array}\right|}{\sqrt{\left(\frac{1}{4}\right)^{2}+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{6}\right)^{2}}}=\frac{(1)\left(\frac{1}{12}+\frac{1}{12}\right)+2\left(\frac{1}{6}+\frac{1}{12}\right)+\left(1-\frac{1}{2}\right) \mid}{\sqrt{\frac{1}{16}+\frac{1}{4}+\frac{1}{36}}}\)
\(=\frac{\left|\frac{1}{6}+\frac{1}{2}+\frac{1}{2}\right|}{\sqrt{\frac{9+36+4}{144}}}=\frac{\left|\frac{7}{6}\right|}{\sqrt{\frac{49}{144}}}=\frac{7}{6} \times \frac{12}{7}=2\)