Let $w=(1,2,3,1)$ be a vector in $\mathbb{R^4}$ find an orthogonal basis for $W^{\perp}$
So of course we can use Gram-Schmidt process, but I would like to try it without, If we find The null space of $\langle(1,2,3,1),(a,b,c,d)\rangle=0$ we will get a basis which is not orthogonal, can we create more terms on $a,b,c,d$ to make the basis orthogonal?
We are looking for three orthogonal vectors $\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$ such that $$a+2b+3c+d = 0.\tag 1$$ We can find two of them very easily by ensuring that $c,d=0$ in the first vector and $a,b=0$ in the second vector. For example, $v_1 = \begin{bmatrix}-2\\1\\0\\0\end{bmatrix}$ and $v_2 = \begin{bmatrix}0\\0\\1\\-3\end{bmatrix}$ lie in $W^{\perp}$. For the third vector, you may find a vector that satisfies $(1)$ then remove its projections along $v_1$ and $v_2$.
I would start as in Math Lover's answer. Then, I want to have a fourth vector which is orthogonal to$$\begin{bmatrix}1\\2\\3\\1\end{bmatrix}\text{, to }\begin{bmatrix}-2\\1\\0\\0\end{bmatrix}\text{, and to }\begin{bmatrix}0\\0\\1\\-3\end{bmatrix}.$$In order to find such a vector, I would use the $4$-dimensional cross-product.