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?