Find $k \in \mathbb{R}$ such that $\mathbb{R}^3=S \oplus T$, with $S=\operatorname{span}\left\{\left[\begin{array}{c}-3 \\ 4 \\ 1\end{array}\right]\right\}$ and $T=\left\{(x, y, z) \in \mathbb{R}^3: k x+y-z=0\right\}.$
Two subspaces $S$ and $T$ are complementary if their intersection is the zero vector space, and their direct sum spans the entire space.
So, the intersection can be found by making S and T equal
so we get
\begin{aligned} -3 a & =k x \\ 4 a & =y \\ a & =-z \end{aligned}
Then, $a = -\frac{3}{k}x$, we set $a+z$ and get that $k$ should not be $0$ right?
Note that if $S=\text{span}\left\{\begin{bmatrix} -3\\ 4\\ 1 \end{bmatrix}\right\}$, then $S=\left\{a\begin{bmatrix} -3\\ 4\\ 1 \end{bmatrix}:a\in\mathbb{R}\right\}$. This vector space is isomorphic to $\tilde{S}=\left\{a(-3,4,1):a\in\mathbb{R}\right\}$.
Also, note that $T=\left\{(x,y,z)\in\mathbb{R}^3:kx+y-z=0\right\}\Leftrightarrow T=\{(x,y,kx+y):x,y\in\mathbb{R}\}$.
Since $\mathbb{R}^3=\tilde{S}\oplus T$, we must have $\tilde{S}\cap T=\{(0,0,0)\}$. This occurs for any real number $k\neq 1$ (why?).