P is the y-intercept of the line SP. To find the y-intercept, set x = 0 in the equation x + ay − a = 0.

To find the y-intercept, set $x = 0$ in the equation $x + ay - a = 0$.

$$0 + ay - a = 0$$ $$ay = a$$

$$y = 1$$

The coordinates of P are $\boxed{(0, 1)}$.

To find the x-intercept, set $y = 0$ in the equation $x + ay - a = 0$.

$$x + a(0) - a = 0$$ $$x - a = 0$$ $$x = a$$

So, the coordinates of S are $(a, 0)$.

The length $OS$ is the distance from the origin $(0,0)$ to $S(a,0)$, which is $a$ (since $a>0$).
The length $OP$ is the distance from the origin $(0,0)$ to $P(0,1)$, which is 1.
Substitute these values into the relation:

$$a = 3(1)$$ $$a = 3$$

The value of $a$ is $\boxed{3}$.

Using $a=3$, the equation of SP is $x + 3y - 3 = 0$.
Rearrange the equation to the form $y = mx + c$:

$$3y = -x + 3$$ $$y = -\frac{1}{3}x + 1$$

The gradient of PS is $m_{PS} = -\frac{1}{3}$.

For perpendicular lines, the product of their gradients is -1.

$$m_{RT} \times m_{PS} = -1$$ $$m_{RT} \times \left(-\frac{1}{3}\right) = -1$$ $$m_{RT} = 3$$

This means the y-intercept $c$ for line RT is $-\frac{5}{3}$.
Substitute the gradient $m_{RT} = 3$ and the y-intercept $c = -\frac{5}{3}$ into $y = mx + c$.

$$y = 3x - \frac{5}{3}$$

The equation of RT is $\boxed{y = 3x - \frac{5}{3}}$.

We have their equations:
Line PS: $y = -\frac{1}{3}x + 1$
Line RT: $y = 3x - \frac{5}{3}$
Set the $y$ values equal to find the $x$-coordinate of R:

$$-\frac{1}{3}x + 1 = 3x - \frac{5}{3}$$

$$-x + 3 = 9x - 5$$ $$3 + 5 = 9x + x$$ $$8 = 10x$$ $$x = \frac{8}{10}$$ $$x = \frac{4}{5}$$

Using $y = 3x - \frac{5}{3}$:

$$y = 3\left(\frac{4}{5}\right) - \frac{5}{3}$$ $$y = \frac{12}{5} - \frac{5}{3}$$

Find a common denominator, which is 15:

$$y = \frac{12 \times 3}{5 \times 3} - \frac{5 \times 5}{3 \times 5}$$ $$y = \frac{36}{15} - \frac{25}{15}$$ $$y = \frac{11}{15}$$

The coordinates of R are $\boxed{\left(\frac{4}{5}, \frac{11}{15}\right)}$.