Find the value of the expression below

If $pqr = 19(p + q + r)$ with $p,q,r$ all prime then by the Fundamental Theorem of Arithmetic we know that one of $p,q,r$ must equal $19$ . So without loss of generality let $p = 19$ . The equation then becomes

$qr = 19 + q + r \Longrightarrow qr - q - r = 19 \Longrightarrow (q - 1)(r - 1) = 20$ .

Now since $q,r \gt 1$ we need to consider the positive factorizations of $20$ into pairs of factors and then assign these pairs to $q -1$ and $r - 1$ . The possible pairs, (without concern for order), are $(1,20), (2,10)$ and $(4,5)$ , which lead to respective values for $(q,r)$ of $(2,21), (3,11)$ and $(5,6)$ . The only one of these pars where both $q,r$ are prime is $(3,11)$ and thus

$p^{2} + q^{2} + r^{2} = 19^{2} + 3^{2} + 11^{2} = \boxed{491}$ .