Solution
= Solution
Eliminating $2y-z$ from the two <plane> equations gives $x=2$, followed by $2y-z=2$. Taking $y=1$ gives the point $\boxed{p=(2,1,0)}$. The entire intersection <line> is $\mathbf r=p+tu$, where $u=(0,1,2)$.
= Solution
Eliminating $2y-z$ from the two <plane> equations gives $x=2$, followed by $2y-z=2$. Taking $y=1$ gives the point $\boxed{p=(2,1,0)}$. The entire intersection <line> is $\mathbf r=p+tu$, where $u=(0,1,2)$.