Write the plane through tfracx-32=tfracy+41=(z-2)/(-3)…

Write the plane through $\tfrac{x-3}2=\tfrac{y+4}1=\tfrac{z-2}{-3}$ parallel to $\tfrac{x+5}4=\tfrac{y-2}7=\tfrac{z-1}2$, and find the shortest distance between the lines.

Explanation of Method

The plane through one line parallel to another, and the shortest distance between the two lines.

Step 1: Plane

\[ \vec n=\{2,1,-3\}\times\{4,7,2\}=\{23,-16,10\},\qquad M_1(3,-4,2) . \]

\[23x-16y+10z-153=0\]

Step 2: Distance

The second line is parallel to this plane, so the distance between the lines equals the distance of $M_2(-5,2,1)$ from the plane: $d=\frac{|-115-32+10-153|}{\sqrt{885}}=\frac{290}{\sqrt{885}}$.

\[d=\frac{290}{\sqrt{885}}=\frac{58\sqrt{885}}{177}\approx9.75\]

This is the same value as the triple-product formula: $\overrightarrow{M_1M_2}\cdot\vec n=-290$.

$d$ $\ell_1$ $\ell_2$

The plane through $\ell_1$ parallel to $\ell_2$; the distance of $\ell_2$ from this plane (dashed) is the shortest distance $\tfrac{290}{\sqrt{885}}\approx9.75$.

Result

\[\boxed{\,23x-16y+10z-153=0\,}\]

\[\boxed{\,d=\frac{290}{\sqrt{885}}=\frac{58\sqrt{885}}{177}\approx9.75\,}\]