Solution:
It is known to us that the shortest distance between two lines $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$ is given as:
$\mathrm{d}=\frac{\left|\begin{array}{ccc}
\mathrm{x}_{2}-\mathrm{x}_{1} & \mathrm{y}_{2}-\mathrm{y}_{1} & \mathrm{z}_{2}-\mathrm{z}_{1} \\
\mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\
\mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2}
\end{array}\right|}{\sqrt{\left(\mathrm{b}_{1} \mathrm{c}_{2}-\mathrm{b}_{2} \mathrm{c}_{1}\right)^{2}+\left(\mathrm{c}_{1} \mathrm{a}_{2}-\mathrm{c}_{2} \mathrm{a}_{1}\right)^{2}+\left(\mathrm{a}_{1} \mathrm{~b}_{2}-\mathrm{a}_{2} \mathrm{~b}_{1}\right)^{2}}}$
The standard form of a pair of Cartesian lines is:
$\frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{a}_{1}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{~b}_{1}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{c}_{1}} \text { and } \frac{\mathrm{x}-\mathrm{x}_{2}}{\mathrm{a}_{2}}=\frac{\mathrm{y}-\mathrm{y}_{2}}{\mathrm{~b}_{2}}=\frac{\mathrm{z}-\mathrm{z}_{2}}{\mathrm{c}_{2}}$
The given equations are:
$\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} \text { and } \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$
On comparing the given equations with the standard form we obtgain,
$\begin{array}{l}
\mathrm{x}_{1}=-1, \mathrm{y}_{1}=-1, \mathrm{z}_{1}=-1 \\
\mathrm{x}_{2}=3, \mathrm{y}_{2}=5, \mathrm{z}_{2}=7 \\
\mathrm{a}_{1}=7, \mathrm{~b}_{1}=-6, \mathrm{c}_{1}=1 \\
\mathrm{a}_{2}=1, \mathrm{~b}_{2}=-2, \mathrm{c}_{2}=1
\end{array}$
Considering
$\left|\begin{array}{ccc}
\mathrm{x}_{2}-\mathrm{x}_{1} & \mathrm{y}_{2}-\mathrm{y}_{1} & \mathrm{z}_{2}-\mathrm{z}_{1} \\
\mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\
\mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2}
\end{array}\right|=\left|\begin{array}{ccc}
3-(-1) & 5-(-1) & 7-(-1) \\
7 & -6 & 1 \\
1 & -2 & 1
\end{array}\right|=\left|\begin{array}{ccc}
3+1 & 5+1 & 7+1 \\
7 & -6 & 1 \\
1 & -2 & 1
\end{array}\right|$
$\begin{array}{l}
=\left|\begin{array}{ccc}
4 & 6 & 8 \\
7 & -6 & 1 \\
1 & -2 & 1
\end{array}\right| \\
=4(-6+2)-6(7-1)+8(-14+6) \\
=4(4)-6(6)+8(-8) \\
=-16-36-64 \\
=-116
\end{array}$
Now, consider
$\begin{array}{l}
\sqrt{\left(b_{1} c_{2}-b_{2} c_{1}\right)^{2}+\left(c_{1} a_{2}-c_{2} a_{1}\right)^{2}+\left(a_{1} b_{2}-a_{2} b_{1}\right)^{2}} \\
=\sqrt{((-6 \times 1)-(-2 \times 1))^{2}+((1 \times 1)-(1 \times 7))^{2}+((7 \times-2)-(1 \times-6))^{2}} \\
=\sqrt{(-6+2)^{2}+(1-7)^{2}+(-14+6)^{2}}=\sqrt{(-4)^{2}+(-6)^{2}+(-8)^{2}} \\
=\sqrt{16+36+64}=\sqrt{116}
\end{array}$
Substituting all the values in eq. (1), we obtain
The shortest distance between the two lines,
$\mathrm{d}=\left|\frac{-116}{\sqrt{116}}\right|=\frac{116}{\sqrt{116}}=\sqrt{116}=2 \sqrt{29}$
As a result, the shortest distance is $2 \sqrt{29}$