Find the orthogonal bases of the space $ V $ and $ V^{\perp}$ In space $\mathbb R^3 $ f Find the orthogonal bases of the space $ V $ and $ V^{\perp}$ where $$V = \left\{ \vec{x} \in \mathbb R^3 : x_1 - 3x_2 + x_3 = 0 \right\} $$
On the begining, I know that may I ask you for basics of topic, but I truly have some doubts... 
Ok, I found basis of $V$: $ [3,1,0]^T,[-1,0,1]^T$  and I have used Gram–Schmidt process to find orthogonal basis of $V$: 
$[3,1,0]^T,[-1/10,3/10,1]^T$ 
Ok, now I should find basis of $ V^{\perp} $ and on the same algorithm find orthogonal basis of $V^{\perp}$ - but there is my question - if in basis of $V^{\perp}$ I have all vectors which are perpendicular to vectors from basis $V$, the basis of $ V^{\perp} $ will be just orthogonal basis of $V$? And orthogonal basis of $V^{\perp}$ will be just basis of $V$? Have I right or there is something other to do?

You have a mistake in your Gram-Schmidt, it should be:
$$\left\{\frac1{\sqrt{10}}\begin{bmatrix} 3 \\ 1 \\ 0\end{bmatrix}, \frac1{\sqrt{110}}\begin{bmatrix} -1 \\ 3 \\ 10\end{bmatrix}\right\}$$
To find the orthonormal basis for $V^\perp$, notice that $\dim V^\perp = \dim\mathbb{R}^3 - \dim V = 3-2 = 1$ so it suffices to find one vector perpendicular to $V$ and normalize it:
$$\frac1{\sqrt{11}}\begin{bmatrix} 1 \\ -3 \\ 1\end{bmatrix}$$
