The numbers expressible as a product of three primes are each of the form $p^{3}, p^{2} q$, or $p q r$, where $p, q$, and $r$ are distinct primes. Now, $\phi\left(p^{3}\right)=p^{2}(p-1), \phi\left(p^{2} q\right)=$ $p(p-1)(q-1)$, and $\phi(p q r)=(p-1)(q-1)(r-1)$. We require $11^{3}+1=12 \cdot 111=2^{2} 3^{2} 37$. The first case is easy to rule out, since necessarily $p=2$ or $p=3$, which both fail. The second case requires $p=2, p=3$, or $p=37$. These give $q=667,223$, and 2, respectively. As $667=23 \cdot 29$, we reject $2^{2} \cdot 667$, but $3^{2} 233=2007$ and $37^{2} 2=2738$. In the third case, exactly one of the primes is 2, since all other primes are odd. So say $p=2$. There are three possibilities for $(q, r):\left(2 \cdot 1+1,2 \cdot 3^{2} \cdot 37+1\right),(2 \cdot 3+1,2 \cdot 3 \cdot 37+1)$, and $\left(2 \cdot 3^{2}+1,2 \cdot 37+1\right)$. Those are $(3,667),(7,223)$, and $(19,75)$, respectively, of which only $(7,223)$ is a pair of primes. So the third and final possibility is $2 \cdot 7 \cdot 223=3122$.