Step 1: Statement I via a transformation matrix. Write the new vectors as $\alpha+\beta,\ \beta+\gamma,\ \gamma+\alpha$. In terms of the basis $\{\alpha,\beta,\gamma\}$ (restricted to their span), the coefficient matrix taking $(\alpha,\beta,\gamma)$ to the new triple is $$M = \begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}$$ (rows give the coefficients of $\alpha,\beta,\gamma$ in each new vector).
Step 2: Compute $\det(M) = 1(1\cdot1 - 0\cdot1) - 0(1\cdot1-0\cdot0) + 1(1\cdot1-1\cdot0) = 1 - 0 + 1 = 2 \neq 0$.
Step 3: Since $M$ is invertible and $\alpha,\beta,\gamma$ are linearly independent, the transformed set $\alpha+\beta,\beta+\gamma,\gamma+\alpha$ must also be linearly independent (an invertible linear transformation preserves linear independence). Statement I is correct.
Step 4: Statement II via a determinant minor. Take the first three coordinates (coefficients of $x^3,x^2,x$) of the three polynomials: $u_1=(2,1,1)$, $u_2=(1,3,1)$, $u_3=(1,2,-1)$.
Step 5: Compute the determinant $$\begin{vmatrix} 2 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & -1 \end{vmatrix} = 2(3(-1)-1\cdot2) - 1(1(-1)-1\cdot1) + 1(1\cdot2-3\cdot1) = 2(-5) - 1(-2) + 1(-1) = -10+2-1 = -9$$
Step 6: Since this $3\times 3$ minor is nonzero, the three original 4-dimensional vectors have rank 3, i.e. they are linearly independent. Statement II is correct.
\[\boxed{\text{Both I and II are correct}}\]