We are going to show that this can only happen when
$$\angle CBP = \angle BCQ = 15^{\circ}.$$

Lemma. If $\angle CBP > \angle BCQ$, then $RT > ST$.

Proof. Let $AD$, $BE$ and $CF$ be the altitudes of triangle $ABC$ concurrent at its centre $G$. Then $P$ lies on $CE$, $Q$ lies on $BF$, and thus $T$ lies in triangle $BDG$.

Note that -
$$\angle FAS = \angle FCQ = 30^{\circ} - \angle BCQ > 30^{\circ} - \angle CBP = \angle EBP = \angle EAR.$$
Since $AF = AE$, we have $FS > ER$ so that
$$GS = GF - FS < GE - ER = GR.$$
Let $T_x$ be the projection of $T$ onto $BC$ and $T_y$ be the projection of $T$ onto $AD$, and similarly for $R$ and $S$. We have
$$R_x T_x = DR_x + DT_x > |DS_x - DT_x| = S_x T_x$$
and
$$R_y T_y = GR_y + GT_y > GS_y + GT_y = S_y T_y.$$
It follows that $RT > ST$.

[1 mark for stating the Lemma, 3 marks for proving it.]

Thus, if $\triangle TRS$ is equilateral, we must have $\angle CBP = \angle BCQ$.

It is clear from the symmetry of the figure that $TR = TS$, so $\triangle TRS$ is equilateral if and only if $\angle RTA = 30^{\circ}$. Now, as $BR$ is an altitude of the triangle $ABC$, $\angle RBA = 30^{\circ}$. So $\triangle TRS$ is equilateral if and only if $RTBA$ is a cyclic quadrilateral. Therefore, $\triangle TRS$ is equilateral if and only if $\angle TBR = \angle TAR$. But
$$\begin{aligned} 90^{\circ} &= \angle TBA + \angle BAR \\ &= (\angle TBR + \angle RBA) + (\angle BAT + \angle TAR) \\ &= (\angle TBR + 30^{\circ}) + (30^{\circ} + \angle TAR) \end{aligned}$$
and so
$$30^{\circ} = \angle TAR + \angle TBR.$$
But these angles must be equal, so $\angle TAR = \angle TBR = 15^{\circ}$. Therefore $\angle CBP = \angle BCQ = 15^{\circ}$.

[3 marks for finishing the proof with the assumption that $\angle CBP = \angle BCQ$.]