Is properly quantified 3SAT complete for PSPACE and all PH levels?

Question

I know 3SAT is NP-complete and QSAT is PSPACE-complete. However, is it true that

$$\exists X_1 \forall X_2 \cdots Q_k X_k \colon \varphi(X_1, \ldots, X_k)$$

is complete for $\Sigma_k$, the existential side of level $k$ of the polynomial hierarchy if each clause in $\varphi$ has at most 3 literals (or some other constant size bound)? Similarly for

$$\forall X_1 \exists X_2 \cdots Q_k X_k \colon \varphi(X_1, \ldots, X_k)$$

in relation to $\Pi_k$, and the quantified version with unbounded alternation in relation to PSPACE, once again with each clause of $\varphi$ having at most 3 literals (or some other constant number)?

Answer 1

QSAT with only 2 literals per clause was shown to be in P back in 1979.

Since the $\Sigma_k$ and $\Pi_k$ versions of QSAT can be reduced to equisatisfiable problems with 3 literals per clause using only a slight modification the method used to transform a SAT instance to 3SAT (i.e. putting all the new variables in the innermost existential group), those limited problems continue to be complete for their levels of the polynomial hierarchy.