= Solution
For the <special linear group> $G=\mathrm{SL}_3(\overline{\mathbb F}_q)$, let $N=q-1$, and choose a generator $\chi$ of the <character group> of $\mathbb F_q^\times$. Every <linear character> of the diagonal <rational maximal torus> has a unique expression
$$
\theta\bigl(\operatorname{diag}(a,b,(ab)^{-1})\bigr)=\chi(a)^r\chi(b)^s,\qquad r,s\in\mathbb Z/N\mathbb Z.
$$
The upper triangular <Borel subgroup> $B=TU$ is rational. Consequently its <Deligne-Lusztig induction> is the actual <principal series of a finite reductive group>
$$
R_T^G(\theta)=\operatorname{Ind}_{B^F}^{G^F}\widetilde\theta,
$$
where $\widetilde\theta$ is the <inflation of a group representation> across $B^F\to T^F$. The geometric reason for this split case is $X(1)=G^F/B^F$: the associated torus covering supplies $\theta$, and the auxiliary unipotent fibres contribute a single even cohomological degree, hence no minus sign.
The rational <Bruhat decomposition> indexes $B^F\backslash G^F/B^F$ by $W=S_3$. Applying the <Mackey restriction formula> and <Frobenius reciprocity> gives
$$
\left\langle R_T^G(\theta),R_T^G(\theta)\right\rangle_{G^F}=\sum_{w\in S_3}\dim\operatorname{Hom}_{B^F\cap{}^{\dot w}B^F}\bigl(\widetilde\theta,{}^{\dot w}\widetilde\theta\bigr)=|\operatorname{Stab}_{S_3}(\theta)|.
$$
Each intersection contains $T^F$, and the two <linear characters> are trivial on its unipotent part. Its contribution is therefore one exactly when $w$ fixes $\theta$, and zero otherwise. Complete reducibility over $\mathbb C$ makes this <character inner product> the sum of squares of <irreducible representation> multiplicities. Thus the <principal series of a finite reductive group> is irreducible exactly when the <Weyl group> stabilizer is trivial.
To compute that stabilizer, represent $\theta$ by the triple $(\chi^r,\chi^s,1)$. Two triples give the same <linear character> on determinant-one diagonal matrices exactly when they differ by a common factor $(\nu,\nu,\nu)$. For example, triviality of $(\eta_1,\eta_2,\eta_3)$ on all such matrices first gives $\eta_1=\eta_3$ by varying $a$, and then $\eta_2=\eta_3$ by varying $b$. The <Weyl group> permutes the three entries.
A transposition fixes $\theta$ precisely when its two interchanged entries coincide: the third, fixed entry forces the common factor to be one. These three obstructions are $r=0$, $s=0$, and $r=s$. A three-cycle can also fix $\theta$ without any two entries coinciding. Its common factor must satisfy $\nu^3=1$; in the nontrivial case the entries, up to order and common factor, are $(1,\nu,\nu^2)$. This happens exactly when $3\mid N$ and $\{r,s\}=\{N/3,2N/3\}$.
The complete condition is therefore
$$
\boxed{r\not\equiv0,\quad s\not\equiv0,\quad r-s\not\equiv0\pmod N,\qquad 3\mid N\Longrightarrow\{r,s\}\ne\{N/3,2N/3\}.}
$$
In particular \b[pairwise distinct coordinate characters alone are insufficient]. For $q=4$ the only distinct triples have the order-three symmetry, and no character of this split <rational maximal torus> gives an irreducible <principal series of a finite reductive group>.
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