Write the plane perpendicular to 2x+3y-z=4 and passing…

Write the plane perpendicular to $2x+3y-z=4$ and passing through $\tfrac{x-1}1=\tfrac{y+1}2=\tfrac{z+2}2$.

Step 1: Solution

The normal is perpendicular to $\vec d=\{1,2,2\}$ and to $\vec n_1=\{2,3,-1\}$: $\vec n=\vec d\times\vec n_1=\{-8,5,-1\}$.

\[8x-5y+z-11=0\qquad(\text{equivalently }-8x+5y-z+11=0)\]

Check: $A(1,-1,-2)$ gives $8+5-2-11=0$; $\vec n\cdot\vec d=-8+10-2=0$; $\vec n\cdot\vec n_1=-16+15+1=0$.