Find the matrix B such that it's null space is $V=\operatorname{span}\{(1,1,0),~(1,2,0),~(1,5,0)\}$.
Since this set of vectors is linearly dependent, I take $\operatorname{span}((1,1,0),~(1,2,0))$ as the linearly independent set. Then, if I think about special solutions, I'd have two, and one pivot variable. Since $(1,1,0)$ is the solution you get by considering $x_1=0,x_2=1$ and $(1,2,0)$ taking $x_1=1,x_2=0$, then the column with the only pivot of the matrix is the third one, $B^3=(0,0,1)^T$
However, I can't figure out how to complete the $3\times 3$ matrix using these special solutions.
Best Answer
The span of those three vectors is $S=\{(x,y,0)\mid x,y\in\Bbb R\}$. So, you can take $B=\left[\begin{smallmatrix}0&0&1\end{smallmatrix}\right]$. It is clear that\begin{align}B.(x,y,z)=0&\iff z=0\\&\iff(x,y,z)\in S.\end{align}