Let $P(t_1)$ and $Q(t_2)$
The slope of PQ is $$\frac{2}{t_1+t_2}=-\frac lm$$
Also $R(\frac{-1}{t_1})$ and $T(\frac{-1}{t_2})$
Slope of $RT$ $$\frac{-2}{\frac{1}{t_1}+\frac{1}{t_2}}$$
But I can’t eliminate the $t_1t_2$ term by simply substituting the values. The answer is $-\frac 1m$
Let $P(\frac{t_1^2}{4a}, t_1)$ and $Q(\frac{t_2^2}{4a}, t_2)$. Substitute $lx+my=1$ into $y^2=4ax$,
$$y^2+\frac{4am}ly-\frac{4a}l=0$$
which leads to,
$$t_1+t_2 = -\frac{4am}l,\>\>\>\>\>t_1t_2 = -\frac {4a}l\tag 1$$
With the focus at $(a,0)$, the points $R(\frac{t_R^2}{4a}, t_R)$ and $T(\frac{t_T^2}{4a}, t_T)$ respectively satisfy
$$\frac{t_R-0}{\frac{t_R^2}{4a}-a} = \frac{t_1-0}{\frac{t_1^2}{4a}-a}, \>\>\>\>\>\>\>\frac{t_T-0}{\frac{t_T^2}{4a}-a} = \frac{t_2-0}{\frac{t_2^2}{4a}-a}$$ which yields
$$t_R = -\frac{4a^2}{t_1},\>\>\>\>\>t_T = -\frac{4a^2}{t_2}$$ Then, the slope of the chord $RT$ is
$$ {Slope}_{RT}=\frac{t_T-t_R}{\frac{t_T^2}{4a} - \frac{t_R^2}{4a}} =- \frac{t_1t_2}{a(t_1+t_1)}=-\frac1{ma}$$
where the results (1) are used in the last step.