Write the plane through (x-3)/(-2)=tfracy+11=tfrac z2…

Write the plane through $\tfrac{x-3}{-2}=\tfrac{y+1}1=\tfrac z2$ parallel to $\tfrac{x+1}1=\tfrac{y-2}3=\tfrac{z-1}2$.

Step 1: Solution

\[ \vec n=\{-2,1,2\}\times\{1,3,2\}=\{-4,6,-7\},\qquad A(3,-1,0) . \]

\[4x-6y+7z-18=0\]

Check: $A$ gives $12+6-18=0$; $\vec n$ is perpendicular to both directions. The second line is not in the plane: its point $(-1,2,1)$ gives $-4-12+7-18\ne0$.