I have the subspaces:
$$S = \langle(2,1,-1), (-1,2,0)\rangle, \qquad T = \{ X + Y + 2Z =0; X - Y - Z = 0\}.$$
I got that $T = \langle(1, 3,-2)\rangle$.
All vectors are linearly independent, so $S + T = \mathbb{R}^3.$ Then I tried to calculate the orthogonal complement for both.
It is my understanding that the orthogonal complement for $S$ should give a vector that is a multiple of $(1,3,-2)$ (generates the same subspace as $T$) and the orthogonal complement for $T$ should give vectors that are linearly dependent from the ones in $S$ (generates the same subspace).
However, complement of $S = t(2,1,3)$ from my calculations, and complement of $T = t(-3,1,0) + s(2,0,1)$.
What I did was resolving the system
$$2x + y - z = 0 \\ - x + 2y = 0$$
For $S$, and for $T$:
$$x + 3y - 2z = 0 .$$
Am I doing something wrong?
Edit: Maybe I'm understanding wrong a part of the theory, but I read that $S$ and its orthogonal complement will generate $\mathbb{R}^n$. The orthogonal complement for T is indeed resulting in $0$ for the scalar product. But if $\dim(S) + \dim(S^\bot) = 3$, and $\dim(S) + \dim(T) = 3$, shouldn't $T$ and $S^\bot$ be the same subspace?
There is absolutely no reason why a vector orthogonal to $S$ should be an element of $T$. This would be true if $T$ were actually the orthogonal complement of $S$, but it is not. Clearly $(1,3,-2)$ is not orthogonal to $S$.
You seem to believe that a complement of $S$ is unique. It is not the case (take the line generated by any vector outside $S$).
PS. Note that your calculations for the complement of $S$ are certainly wrong, since the vector you obtain is not orthogonal to the generators of $S$...