The shortest distance between the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}\) is

The lines are
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \text { and } \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}\)
Comparing equations with
\(\frac{x-x_1}{\mathrm{a}_1}=\frac{y-y_1}{\mathrm{~b}_1}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{c}_1} \text { and } \frac{x-x_2}{\mathrm{a}_2}=\frac{y-y_2}{\mathrm{~b}_2}=\frac{\mathrm{z}-\mathrm{z}_2}{\mathrm{c}_2} \text {, }\)
we get
\(\begin{aligned}
& x_1=1, y_1=2, \mathrm{z}_1=3 \quad x_2=2, y_2=4, \mathrm{z}_2=5 \\
& \mathrm{a}_1=2, \mathrm{~b}_1=3, \mathrm{c}_1=4 \quad \mathrm{a}_2=3, \mathrm{~b}_2=4, \mathrm{c}_2=5 \\
& \left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
\mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\
\mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2
\end{array}\right|=\left|\begin{array}{ccc}
1 & 2 & 2 \\
2 & 3 & 4 \\
3 & 4 & 5
\end{array}\right| \\
& =1(15-16)-2(10-12)+2(8-9) \\
& =1 \end{aligned}\)