First, resolve the forces perpendicular to the plane to find the normal reaction \(R\):

\(R = 13g \cos 22.6 = 13g \times \frac{12}{13} = 120 \text{ N}\)

Next, calculate the frictional force \(F\) using \(F = \mu R\):

\(F = 0.3 \times 120 = 36 \text{ N}\)

Apply Newton's second law parallel to the plane (since the particle moves at constant speed, acceleration \(a = 0\)):

\(T = F + 13g \sin 22.6 = 36 + 13g \times \frac{5}{13} = 86 \text{ N}\)

The work done \(WD\) by the force \(T\) is given by:

\(WD = T \times d = 86 \times 2.5 = 215 \text{ J}\)