Let the four numbers be $a, b, c, d$. We see that $(a+b)^2+(c+d)^2+(a+c)^2+(b+d)^2+(a+d)^2+(b+c)^2$ $=2(a^2+b^2+c^2+d^2)+(a+b+c+d)^2=2(21)+9^2=123$. So that means that one of the sums $(a+b)^2+(c+d)^2, (a+c)^2+(b+d)^2, (a+d)^2+(b+c)^2$ is at least $\frac{123}{3}=41$. But all three of them are of the form $m^2+(9-m)^2$. If $m^2+(9-m)^2\geq 41$, then expanding gives $2m^2-18m+81\geq 41$, or $m^2-9m+20=(m-4)(m-5)\geq 0$. Thus either $m\leq 4$ or $m\geq 5$. Either way, one of the six sums $a+b, b+c, c+d, a+c, b+d, a+d$ is at least 5. WLOG it's $a+b$. Then $21+2(ab-cd)=a^2+b^2+c^2+d^2+2(ab-cd)=(a+b)^2+(c-d)^2\geq (a+b)^2\geq25$, so that $ab-cd\geq2$. Equality happens when $a+b=5$ and $c-d=0$, so $c=d=2$ and $a^2+b^2=21-c^2-d^2=13$, and hence $a$ and $b$ are $2$ and $3$. So equality happens when $\{a, b, c, d\}=\{2, 2, 2, 3\}$ taken as a multiset.