Triangle PQR has vertices P(3,2),Q(-1,1) and R(-3-1)

Given P$(3,2)$ and Q$(-1,1)$:

$$PQ = \sqrt{(-1-3)^2 + (1-2)^2}$$ $$PQ = \sqrt{(-4)^2 + (-1)^2}$$ $$PQ = \sqrt{16 + 1}$$ $$PQ = \sqrt{17}$$

Given Q$(-1,1)$ and R$(-3,-1)$:

$$QR = \sqrt{(-3-(-1))^2 + (-1-1)^2}$$ $$QR = \sqrt{(-3+1)^2 + (-2)^2}$$ $$QR = \sqrt{(-2)^2 + (-2)^2}$$ $$QR = \sqrt{4 + 4}$$ $$QR = \sqrt{8}$$ $$QR = 2\sqrt{2}$$

Given R$(-3,-1)$ and P$(3,2)$:

$$RP = \sqrt{(3-(-3))^2 + (2-(-1))^2}$$ $$RP = \sqrt{(3+3)^2 + (2+1)^2}$$ $$RP = \sqrt{6^2 + 3^2}$$ $$RP = \sqrt{36 + 9}$$ $$RP = \sqrt{45}$$ $$RP = 3\sqrt{5}$$

The lengths of the sides are:
$PQ = \boxed{\sqrt{17}}$
$QR = \boxed{2\sqrt{2}}$
$RP = \boxed{3\sqrt{5}}$