[solution]
**(a) $S_1$ is a subspace.** Let $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$ satisfy $u_1 + 2u_2 - u_3 = 0$ and $v_1 + 2v_2 - v_3 = 0$. For any $\lambda, \mu \in \mathbb{R}$:
\begin{align*}
(\lambda u_1 + \mu v_1) + 2(\lambda u_2 + \mu v_2) - (\lambda u_3 + \mu v_3) = \lambda(u_1 + 2u_2 - u_3) + \mu(v_1 + 2v_2 - v_3) = 0.
\end{align*}
So $\lambda\mathbf{u} + \mu\mathbf{v} \in S_1$. Geometrically, $S_1$ is the plane through the origin with normal vector $(1, 2, -1)$.
**(b) $S_2$ is not a subspace.** The vector $\mathbf{e}_1 = (1, 0, 0) \in S_2$ since $1 \leq 1$, but $2\mathbf{e}_1 = (2, 0, 0) \notin S_2$ since $4 > 1$. Closure under scalar multiplication fails.
**(c) $S_3$ is not a subspace.** The point $(1, 1, 0)$ satisfies $1 = 1^2$, but $(1, -1, 0)$ also satisfies $1 = (-1)^2$. Their sum is $(2, 0, 0)$, which does not satisfy $2 = 0^2$. Alternatively, $\mathbf{0} = (0,0,0) \in S_3$, but $(1,1,0) \in S_3$ while $2(1,1,0) = (2,2,0) \notin S_3$ since $2 \neq 4$.
**(d) $S_4$ is a subspace.** If $u_1 = 2u_3$ and $v_1 = 2v_3$, then $\lambda u_1 + \mu v_1 = 2(\lambda u_3 + \mu v_3)$, so $\lambda\mathbf{u} + \mu\mathbf{v} \in S_4$. This is a plane through the origin in $\mathbb{R}^3$.
[/solution]