I started by competing the square on the second and third equations:

(2Q-R)^2 = 100-3R^2

(2P+R)^2 = 576-3R^2

Then expressed each P and Q in terms of R:

Q = (R +/- sqrt(100-3R^2))/2

P = (R +/- sqrt(576-3R^2))/2

Then substitute into the first equation:

(R +/- sqrt(100-3R^2))/2 + (R +/- sqrt(576-3R^2))/2 = 13

This simplifies to a quartic in R:

27R^4 + 2028R^2 - 57600 = 0

The one positive real root is R = sqrt[2*sqrt(71761)-169]/3

Normally these problems have a tidier final form. Did I miss something?