Write the common normal of (x-3)/(-2)=tfracy+11=tfrac z2 and…

Write the common normal of $\tfrac{x-3}{-2}=\tfrac{y+1}1=\tfrac z2$ and $\tfrac{x+1}1=\tfrac{y-2}3=\tfrac{z-1}2$.

Explanation of Method

The common normal of two skew lines, with its feet and length.

Step 1: Setup

Take $A=(3-2s,-1+s,2s)$ and $B=(-1+t,2+3t,1+2t)$. Their difference is

\[ \overrightarrow{AB}=\{-4+t+2s,\ 3+3t-s,\ 1+2t-2s\}. \]

Requiring $\overrightarrow{AB}\perp\vec d_1$ and $\overrightarrow{AB}\perp\vec d_2$: \[5t-9s+13=0,\qquad 14t-5s+7=0\ \Longrightarrow\ s=\tfrac{147}{101},\ t=\tfrac{2}{101}.\]

Step 2: Result

\[ A\left(\tfrac{9}{101},\tfrac{46}{101},\tfrac{294}{101}\right),\qquad B\left(-\tfrac{99}{101},\tfrac{208}{101},\tfrac{105}{101}\right),\qquad \overrightarrow{AB}=\tfrac{27}{101}\{-4,6,-7\}. \]

\[\frac{x-\frac{9}{101}}{-4}=\frac{y-\frac{46}{101}}{6}=\frac{z-\frac{294}{101}}{-7},\qquad d=\frac{27}{\sqrt{101}}\approx2.69\]

Check with the triple product: $\overrightarrow{M_1M_2}=\{-4,3,1\}$, $\overrightarrow{M_1M_2}\cdot\{-4,6,-7\}=16+18-7=27$, and $|\vec n|=\sqrt{101}$. So $d=\tfrac{27}{\sqrt{101}}$. $\checkmark$

$A$ $B$ $d$ $\ell_1$ $\ell_2$

The common normal $AB$ (dashed), direction $\{-4,6,-7\}$, length $\tfrac{27}{\sqrt{101}}\approx2.69$.

Result

\[\boxed{\,\frac{x-\frac{9}{101}}{-4}=\frac{y-\frac{46}{101}}{6}=\frac{z-\frac{294}{101}}{-7},\qquad d=\frac{27}{\sqrt{101}}\approx2.69\,}\]