Let $W$ be the intersection of the two planes $\pi_{1}$ and $\pi_{2}$ defined by:
\begin{align} \pi_{1} &=\{(x,y,z) \mid x−y−z=0\} \\ \pi_{2}&=\{(x,y,z)∣x+2y+z=0\}. \end{align}
Find a basis for $W^\perp$.
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First of all compute the solutions of the system
$x-y-z=0$
$x+2y+z=0.$
Show that these are given by $t(1,-2,3)$ with real $t$.
Then you get $W=\{t(1,-2,3): t \in \mathbb R\}.$
Now look for two linearly independent vectors $(a,b,c), (u,v,v)$ such that $(a,b,c) \perp (1,-2,3)$ and $(u,v,w) \perp (1,-2,3)$.
Then $\{(a,b,c), (u,v,v)\}$ is a basis of $W^{\perp}.$