Answer
See below
Work Step by Step
Given: $v=(-3,6,1)$
Asuume $w=(w_1,w_2,w_3)$ is orthogonal to $v$
then $=0\\
-3w_1+6w_2+w_3=0\\
\rightarrow w_1=2w_2+\frac{1}{3}w_3$
Obtain: $w=\{(2t+\frac{1}{3}s,t,s);s,t \in R$
If we take $t=1,s=0 \rightarrow w_1=(2,1,0)$
If we take $t=0,s=3 \rightarrow w_2=(1,0,3)$
Obtain: $\begin{bmatrix}
-3 & 6 & 1\\
2 & 1 & 0\\
1 & 0 & 3
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & 3\\
2 & 1 & 0\\
-3 & 6 & 1\\
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & 3\\
0 & 1 & -6\\
0 & 6 & 10\\
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & 3\\
0 & 1 & -6\\
0 & 0 & -26\\
\end{bmatrix}$
We can notice that there are three independent row vectors.
The basis of R^3 is $\{(-3,6,1);(2,1,0);(1,0,3)\}$
