Solution:

We show that the answer is NO. Suppose, if possible, let $a, b, c$ be three distinct positive real numbers such that $a, b, c, b+c-a, c+a-b, a+b-c$ and $a+b+c$ form a 7-term arithmetic progression in some order. We may assume that $a < b < c$. Then there are only two cases we need to check:

(I) $a+b-c < a < c+a-b < b < c < b+c-a < a+b+c$

and

(II) $a+b-c < a < b < c+a-b < c < b+c-a < a+b+c$.

Case I. Suppose the chain of inequalities $a+b-c < a < c+a-b < b < c < b+c-a < a+b+c$ holds. Let $d$ be the common difference. Thus we see that
$$c = a + b + c - 2d, \quad b = a + b + c - 3d, \quad a = a + b + c - 5d$$
Adding these, we see that $a + b + c = 5d$. But then $a = 0$, contradicting the positivity of $a$.

Case II. Suppose the inequalities $a+b-c < a < b < c+a-b < c < b+c-a < a+b+c$ are true. Again we see that
$$c = a + b + c - 2d, \quad b = a + b + c - 4d, \quad a = a + b + c - 5d$$
We thus obtain $a + b + c = \frac{11}{2} d$. This gives
$$a = \frac{1}{2} d, \quad b = \frac{3}{2} d, \quad c = \frac{7}{2} d$$
Note that $a + b - c = a + b + c - 6d = -\frac{1}{2} d$. However, we also get $a + b - c = \left(\frac{1}{2} + \frac{3}{2} - \frac{7}{2}\right) d = -\frac{3}{2} d$. It follows that $3e = e$ giving $d = 0$. But this is impossible.

Thus there are no three distinct positive real numbers $a, b, c$ such that $a, b, c, b+c-a, c+a-b, a+b-c$ and $a+b+c$ form a 7-term arithmetic progression in some order.