Summary: ``It is well known that any graph admits a crossing-free straight-line drawing in $\Bbb{R}^3$ and that any planar graph admits the same even in $\Bbb R^2$. For a graph $G$ and $d \in \{2, 3\}$, let $\rho^1_d(G)$ denote the smallest number of lines in $\Bbb{R}^d$ whose union contains a crossing-free straight-line drawing of $G.$ For $d = 2,$ the graph $G$ must be planar. Similarly, let $\rho^2_3(G)$ denote the smallest number of planes in $\Bbb{R}^3$ whose union contains a crossing-free straight-line drawing of $G.$ We investigate the complexity of computing these three parameters and obtain the following hardness and algorithmic results. \par $\bullet$ For $d \in \{2, 3\},$ we prove that deciding whether $\rho^1_d(G) \le k$ for a given graph $G$ and integer $k$ is $ \Bbb{R}$-complete. \par $\bullet$ Since NP $\subseteq \Bbb{R},$ deciding $\rho^1_d(G) \le k$ is NP-hard for $d \in \{2, 3\}.$ On the positive side, we show that the problem is fixed-parameter tractable with respect to $k.$ \par $\bullet$ Since $\Bbb{R} \subseteq $ PSPACE, both $\rho^1_2 (G)$ and $\rho^1_3(G)$ are computable in polynomial space. On the negative side, we show that drawings that are optimal with respect to $\rho^1_2$ or $\rho^1_3$ sometimes require irrational coordinates. \par $\bullet$ We prove that deciding whether $\rho^2_3(G) \le k$ is NP-hard for any fixed $k \ge 2.$ Hence, the problem is not fixed-parameter tractable with respect to $k$ unless P = NP.''