I have been able to find the answer via exhaustion. I defined $a\le b\le c\le d\le e\le f$ as the primes and solved an inequality for each variable. The answer is $2,2,3,3,7,67$. However, the number of cases to solve got quite large and I was wondering if there is any neat solution to this problem.
Start with the observation of J. W. Tanner that two of the primes must be $3,67$. Note that if all of the primes are odd, their product is odd, but their sum is even, so we can conclude that one of the primes must be $2$.
Restate the problem as $2\cdot 3\cdot 67\cdot p_1p_2p_3=201(p_1+p_2+p_3+72)$ from which $2\cdot p_1p_2p_3=(p_1+p_2+p_3+72)$. Once again, if all unidentified primes are odd then LHS is even but RHS is odd, so there is another factor of $2$ present. This yields $4\cdot p_1p_2=(p_1+p_2+74)$. We can quickly convince ourselves that $p_1\ne p_2$ and $p_1\equiv p_2 \bmod 4$.
The smallest pair of primes that are $\equiv 1 \bmod 4$ are $5,13$, but $4\cdot 5\cdot 13>5+13+74$ and looking at larger primes only makes the inequality greater. The smallest pair of primes that are $\equiv 3 \bmod 4$ are $3,7$, and this provides a solution. The next larger pair of primes that are $\equiv 3 \bmod 4$ are $7,11$, and also in this case, $4\cdot 7\cdot 11>7+11+74$ and looking at larger primes only makes the inequality greater. So the remaining primes are $3,7$.
The complete list is as OP found: $2,2,3,3,7,67$.