The correct answer is: $\boxed{\text{A. }p – q + r = 0 \text{ or } p = q = -r}$.
To solve this, we can use the following theorem:
If a system of equations has a non-trivial solution, then the determinant of the coefficient matrix is equal to zero.
In this case, the coefficient matrix is:
$$\begin{bmatrix}
p & q & r \
q & r & p \
r & p & q
\end{bmatrix}$$
The determinant of this matrix is:
$$p^2 + q^2 + r^2 – 2pq – 2pr – 2qr$$
If this determinant is equal to zero, then the system of equations has a non-trivial solution.
We can factor the determinant as follows:
$$p^2 + q^2 + r^2 – 2pq – 2pr – 2qr = (p – q + r)(p + q – r)(p + q + r)$$
Therefore, the system of equations has a non-trivial solution if and only if one of the following three equations is true:
$$p – q + r = 0$$
$$p + q – r = 0$$
$$p + q + r = 0$$
The first equation is equivalent to $p = q = -r$. The second equation is equivalent to $p = -q = r$. The third equation is equivalent to $p = q = r$.
Therefore, the correct answer is $\boxed{\text{A. }p – q + r = 0 \text{ or } p = q = -r}$.