Primes and Symmetry

Since the sum of the given primes is even, one of them must be $2$ .

What should be noticed is that the expression to be evaluated resembles the Elementary Symmetric Polynomials and hence the Vieta's Formulas .

That said, let $P(x)$ a polynomial whose roots are $p,q,r$ .

Using the Vieta's Formula we can construct that polynomial as

$P(x)=x^3 - (p+q+r)x^2 + (pq + qr + rp)x - pqr$

Since $2$ is one of the roots, we have $P(2)=0$ .

If we substitute $x=2$ ,

$P(2) = 8 - (p+q+r)4 + (pq + qr + rp)2+ pqr=0$

Thus,

$pqr - 2(pq + qr + rp) + 4(p+ q + r) = \boxed{8}$